Module & More Add Math Tg 4

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik

(a) Diberi –1 < h < 1. Iaitu, h = –1, 0, 1 1→2

Suatu graf fungsi f(x) = ax2 + bx + c mempunyai 7q . 1 – 7q , 1
6 2h 6 2h
titik minimum (bentuk graf ) jika a . 0. 1 6 1 6
2h 7 2h 7
Jadi, h = 1. q . × q , × (– )

BAB 2 Given that –1 < h < 1, that is h = –1, 0, 1 q . 3 q , – 3
A function graph f(x) = ax2 + bx + c has minimum 7h 7h

point (graph ) if a . 0. Then, h = 1.

(b) Graf fungsi g(x) = x2 + (k − 1)x + 4 tidak

menyilang paksi-x. Jadi, graf itu tidak

mempunyai punca nyata. 8. Rajah menunjukkan graf bagi fungsi kuadratik
p
The function graph g(x) = x2 + (k – 1)x + 4 does not SPM f(x) = xn + qx + r, dengan keadaan p, q, r, m dan n
intersect the x-axis. Then, the graph does not have 2015

real root.

b2 − 4ac , 0 ialah pemalar.

(k − 1)2 − 4(1)(4) , 0 Diagram shows the graph of a quadratic function
p
k2 − 2k + 1 − 16 , 0 k f(x) = xn + qx + r, where p, q, r, m and n are constants.
k2 − 2k − 15 , 0 5
–3

(k − 5)(k + 3) , 0 f(x)

k − 5 = 0 , k + 3 = 0

k = 5 , k = −3

\ −3 , k , 5

7. Persamaan kuadratik (px)2 + 7qx + 9 = 0 –m 0 mx
SPM mempunyai dua punca yang sama. Manakala,
2019 persamaan kuadratik ℎx2 − 2x + 2p = 0 tiada (a) Nyatakan nilai n.

punca, dengan keadaan ℎ, p dan q ialah pemalar. State the value of n.
Ungkapkan julat q dalam sebutan ℎ.
(b) Jika f(x) = 0 dan hasil darab punca ialah −r,
The quadratic equation (px)2 + 7qx + 9 = 0 has two equal nyatakan nilai
roots. Meanwhile, the quadratic equation ℎx2 − 2x + 2p =
0 has no roots, where ℎ, p and q are constants. Express If f(x) = 0 and the product of roots is −r, state the
the range of q in terms of ℎ. value of

p2x2 + 7qx + 9 = 0 (i) q
a = p2,  b = 7q,  c = 9 (ii) p
b2 – 4ac = 0
(7q)2 – 4(p2)(9) = 0 (a) x–n = x2
–n = 2
49q2 – 362 = 0  n = –2

49q2 = 36p2

√p2 = 49q2
36
(b) f(x) = 0
p = ± 76q …… 1
px2 + qx + r = 0

hx2 – 2x + 2p = 0 a = p,  b = q,  c = r

a = h,  b = –2,  c = 2p (i) Hasil tambah punca / Sum of roots
b2 – 4ac , 0 = m + (–m) = 0
(–2)2 – 4(h)(2p) , 0
4 – 8hp , 0 – b = 0
4 , 8hp a
1 , 2hp
p . 21h …… 2 – ­qp = 0

q = 0

40

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 

(ii) Hasil darab punca / Product of roots = –r (b) Seterusnya, bentukkan persamaan kuadratik
yang mempunyai punca-punca m + 3 dan
c = –r m – 4.
a
Hence, form the quadratic equation with the roots
r = –r m + 3 and m – 4.
p
r = –pr
(a) x2 – 6(2x – h) = 0 BAB 2
r + pr = 0

r(1 + p) = 0 x2 – 12x + 6h = 0

r = 0,  p = –1 a = 1, b = –12, c = 6h

m + 3m = – b
a

9. Graf bagi fungsi kuadratik f(x) = 4[3ℎ − (x − 2)2], 4m = – 1 –12 2
1
SPM dengan keadaan ℎ ialah pemalar mempunyai titik
2019 maksimum (2, ℎ − 11). 4m = 12

The graph of a quadratic function f(x)= 4[3ℎ − (x − 2)2], m = 3

where ℎ is a constant, has maximum point (2, ℎ − 11). m × 3m = c
a
(a) Nyatakan nilai h.
3m2 = 6h
State the value of ℎ. 1

(b) Nyatakan jenis punca bagi f(x) = 0. 3 × 32 = 6h

Justifikasikan jawapan anda. 6h = 27

State the type of roots for f(x) = 0. Justify your h = 27
answer. 6

(a) f(x) = 12h – 4(x – 2)2 9
2
Nilai maksimum / Maximum value = 12h =

h – 11 = 12h

11h = –11 (b) Punca-punca baharu ialah 6 dan –1.

h = –1 The new roots are 6 and –1.

(b) f(x) = 0, –12 – 4(x – 2)2 = 0 HTP baharu/new SOP = 6 + (–1)

–12 – 4(x2 – 4x + 4) = 0 =5

–12 – 4x2 + 16x – 16 = 0 HDP baharu/new POR = 6 × (–1)

–4x2 + 16x – 28 = 0 = –6

–x2 + 4x – 7 = 0 Persamaan baharu ialah

a = –1,  b = 4,  c = –7 The new equation is

b2 – 4ac = 42 – 4(–1)(–7) x2 – (5)x + (–6) = 0

= 16 – 28 = –12 x2 – 5x – 6 = 0

b2 – 4ac , 0

Tidak ada punca nyata / No real roots 2. Persamaan kuadratik k − 4x = x2 − x + 1, dengan

SPM keadaan k ialah pemalar, mempunyai punca-
2015 punca a dan b.

Kertas 2 The quadratic equation k − 4x = x2 − x + 1, where k is a

1. Persamaan kuadratik x2 – 6(2x – h) = 0, dengan constant, has roots a and b.
keadaan h ialah pemalar mempunyai punca-
punca m dan 3m, m ≠ 0. (a) Cari julat nilai k jika a ≠ b.

A quadratic equation x2 – 6(2x – h) = 0, where h is a Find the range of values of k if a ≠ b.

constant has roots m and 3m, m ≠ 0. (b) Diberi a + 1 dan b + 1 adalah punca-punca

(a) Cari nilai m dan nilai h. bagi satu lagi persamaan kuadratik

Find the value of m and of h. 2x2 − hx + 4 = 0, dengan keadaan h ialah

pemalar. Cari nilai k dan nilai h.
Given a + 1 and b + 1 are the roots of another
quadratic equation 2x2 − hx + 4 = 0, where h is a

constant. Find the value of k and of h.

41

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik

(a) x2 − x + 1 = k − 4x 3. Fungi kuadratik f(x) = 2x2 – 8x + k mempunyai

x2 − x + 4x + 1 − k = 0 nilai minimum 5 apabila x = h.

x2 + 3x + 1 − k = 0 The quadratic function f(x) = 2x2 – 8x + k has a minimum
value of 5 when x = h.

(a) Cari nilai h dan nilai k.
BAB 2 a ≠ b bermaksud dua punca berbeza
Find the value of h and of k.
a ≠ b means two different roots
(b) Seterusnya, dengan menggunakan nilai h dan
b2 − 4ac . 0
nilai k di (a), lakarkan graf f(x) = 2x2 – 8x + k.
32 − 4(1)(1 − k) . 0
Hence, by using the value of h and of k in (a), sketch
9 − 4 + 4k . 0 the graph of f(x) = 2x2 – 8x + k.

5 + 4k . 0

4k . −5 (a) f(x) = 2x2 – 8x + k
= 2(x2 – 4x) + k
k . – 5
4
–4 –4
= 23x2 – 4x + 1 2 22 – 1 2 224 + k

(b) Untuk/For 2x2 − hx + 4 = 0 = 2[(x – 2)2 – 4] + k

x2 − h x + 2 = 0 = 2(x – 2)2 – 8 + k
2
–8 + k = 5

HTP/SOR: h = (a + 1) + (b + 1) k = 13
2
x – 2 = 0
h
2 − 2 = a + b …… 1 x = 2

\  h = 2

HDP/POR: 2 = (a + 1)(b + 1)

2 = ab + a + b + 1 …… 2 (b) f(x) = 2(x – 2)2 + 5

  a = 2 > 0

Untuk/For x2 + 3x + 1 − k = 0, f(x) mempunyai nilai minimum
HTP/SOR: −3 = a + b …… 3
HDP/POR: 1 − k = ab …… 4 f(x) has minimum value

Titik minimum ialah (2, 5)

The minimum point is (2, 5)

Gantikan 3 ke 1: h − 2 = −3 f(x) = 2x2 – 8x + 13
Replace 3 into 1 2
h b2 – 4ac = (–8)2 – 4(2)(13)
2
= −1 = –40 < 0

h = −2 Graf f(x) tidak menyilang paksi-x.

Graph f(x) does not intersect the x-axis.

 Gantikan 4 dan 3 ke 2: 2 = (1 − k) + (−3) + 1 Apabila/When x = 0, f(x) = 2(0)2 – 8(0) + 13

  Replace 4 and 3 into 2 2 = −1 − k = 13

k = −3 f(x)

13

(2, 5) x Praktis
0 SPM
Ekstra

42

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 

Sudut KBAT KBAT

Ekstra

1. ABCD ialah sebuah segi empat tepat dengan 2. (a) Cari julat nilai-nilai m dengan keadaan BAB 2
panjang 5x cm dan lebar (4 – x) cm. fungsi f(x) = 2x2 – 7x + m adalah
sentiasa positif bagi semua nilai x.
ABCD is a rectangle with a length of 5x cm and a width
of (4 – x) cm. Find the range of values of m such that the
function f(x) = 2x2 – 7x + m is always positive for
5x all values of x.

4–x (b) Tunjukkan fungsi g(x) = 3x – 8 – 4x2 adalah
sentiasa negatif bagi semua nilai x.
Cari perimeter, dalam cm, segi empat ABCD jika
Show that the function g(x) = 3x – 8 – 4x2 is
luas ABCD adalah maksimum.
always negative for all values of x.
Seterusnya, nyatakan nilai luas yang
(a) f(x) = 2x2 – 7x + m
maksimum, dalam cm2, bagi segi empat ABCD.
a = 2, b = –7, c = m
Find the perimeter, in cm, of the rectangle ABCD if the
area of ABCD is a maximum. b2 – 4ac < 0

Hence, state the maximum value of the area, in cm2, of (–7)2 – 4(2)(m) < 0
the rectangle ABCD.
49 – 8m < 0

Katakan luas segi empat tepat/Let the the area of 8m > 49
rectangle = f(x) 49
m > 8

f(x) = 5x(4 – x) (b) g(x) = – 4x2 + 3x – 8

= –5x2 + 20x a = –4, a < 0, graf maksimum/

= –5(x2 – 4x) 4 4 maximum graph
2 2
= –53x2 – 4x + 1– 22 – 1– 224 b2 – 4ac

= –5(x – 2)2 + 20 = 32 – 4(–4)(–8)

a = –5 < 0, maka, f(x) mempunyai nilai = 9 – 128

maksimum = –119 < 0, tidak mempunyai punca.

a = –5 < 0, hence, f(x) has maximum value does not have roots

x–2=0 g(x) adalah sentiasa negatif.
g(x) is always negative.
\  x = 2
Perimeter = 2(5x) + 2(4 – x) y
Ox
= 2(5 × 2) + 2(4 – 2)
= 24 cm

Luas maksimum = 5(2)(4 – 2)

Maximum area = 20 cm2

Kuiz 2

43

BAB Sistem Persamaan

3 Systems of Equations

3.1 Sistem Persamaan Linear dalam Tiga Pemboleh Ubah

Systems of Linear Equations in Three Variables

NOTA IMBASAN

Sistem persamaan dalam tiga pemboleh ubah
Systems of Linear Equations in Three Variables

NOTA IMBASAN

44

  Matematik Tambahan  Tingkatan 4  Bab 3 Sistem Persamaan 

1. Tentukan sama ada persamaan berikut adalah sistem persamaan linear dalam tiga pemboleh ubah atau bukan.
Determine whether the following equations are systems of linear equations in three variables or not. TP 1

CONTOH (a) a + b + c = 6
2a – b + 3c = 21
2p + 3q – r = 17 a – 3b + c = 2
5p + 2q – r2 = 0
6p – q + 2r = 1 Ya. Semua tiga persamaan mempunyai tiga
pemboleh ubah a, b, dan c dengan kuasa 1.
Bukan. Kuasa bagi r ialah 2. Ini bukan linear.
Yes. All three equations have variables a, b, and c, of
No. The power of r is 2. This is not linear. power 1.

(b) 2m + n – 3p = –12 (c) 2x + y = z BAB 3
3m + 2n – p = –1
5m – n + 6p = 3 3x – 2y – 4z = –2

Ya. Semua tiga persamaan mempunyai tiga 5x – y2 + 2x = 21
pemboleh ubah m, n dan p dengan kuasa 1.
Bukan. Kuasa bagi y ialah 2. Ini bukan linear..
Yes. All three equations have variables m, n and p,
of power 1. No. The power of y is 2. This is not linear.

2. Selesaikan sistem persamaan linear berikut menggunakan kaedah penghapusan.   TP 4

Solve the following systems of linear equations using the elimination method.

CONTOH 1

4p – q + 2r = –3
2p + 3q = 7 – 4r
3p – 2q + r = –7

Penyelesaian:

4p − q + 2r = −3… 1 Susun semula supaya urutan susunan pemboleh
2p + 3q + 4r = 7… 2
3p − 2q + r = −7… 3 ubah itu sama.

Rearrange so that the arrangement of variables is the same.

1 × 3: 12p − 3q + 6r = −9 … 4 Pilih 1 dan 2 untuk menghapuskan q dengan
2p + 3q + 4r = 7 … 2
4 + 2: 14p + 10r = −2 … 5 penambahan.

Choose 1 and 2 to eliminate q by addition.

1 × 2: 8p − 2q + 4r = −6 … 6 Pilih satu pasang persamaan lain untuk
3p − 2q + r = −7 … 3 menghapuskan q.
6 − 3: 5p + 3r = 1 … 7
Choose another pair of equations to eliminate q.

5 × 3: 42p + 30r = −6 … 8 Hapuskan r dalam 5 dan 7

7 × 10: 50p + 30r = 10 … 9 Eliminate r in 5 and 7

9 − 8: 8p = 16

p = 2

5(2)+3r = 1 Gantikan p = 2 ke dalam 7
3r = −9
r = −3 Substitute p = 2 into 7

4(2) − q + 2(−3) = −3 Gantikan p = 2 dan r = –3 ke dalam 1
8 − q − 6 = −3
8 − 6 + 3 = q Substitute p = 2 and r = –3 into 1
q = 5
      \ p = 2, q = 5,  r = −3

45

  Matematik Tambahan  Tingkatan 4  Bab 3 Sistem Persamaan

Selesaikan sistem persamaan berikut menggunakan kaedah penggantian. TP 4

Solve the following systems of linear equations using the substitution method.

CONTOH 2

3x – y + 2z = 25 …… 1
2x + 2y + z = 5 …… 2
4x + 3y – 2z = 2 …… 3
Penyelesaian:

BAB 3 Dari 1 / From 1: Ungkapkan y dalam sebutan x dan z dari 1
3x + 2z – 25 = y
y = 3x + 2z – 25 …… 4 Express y in terms of x and z from 1

Dari 2 / From 2: Gantikan 4 ke dalam 2 dan 3
2x + 2(3x + 2z – 25) + z = 5
2x + 6x + 4z – 50 + z = 5 Substitute 4 into 2 and 3
8x + 5z = 55 …… 5
Dari 3 / From 3:
4x + 3(3x + 2x – 25) – 2z = 2
4x + 9x + 6x – 75 – 2z = 2
13x + 4z = 77 …… 6

5 × 4: 32x + 20z = 220 …… 7 Jadikan pekali z dalam 5 dan 6 sama.
6 × 5: 65x + 20z = 385 …… 8
Make the coefficient of z in 5 and 6 the same.

8 − 7: 33x = 165 Tentukan nilai pekali x, y dan z
x = 5
Dari 5 / From 5: Determine the value of coefficient of x, y and z
8(5) + 5z = 55
5z = 15
z = 3
Dari 4 / From 4:
y = 3(5) + 2(3) – 25
= 15 + 6 – 25
= –4
\  x = 5,  y = –4,  z = 3

46

  Matematik Tambahan  Tingkatan 4  Bab 3 Sistem Persamaan 

Selesaikan sistem persamaan berikut menggunakan kaedah penghapusan atau kaedah penggantian. TP 4

Solve the following system of linear equations using the elimination method or substitution method.

(a) 5x – 2y – 4z = 3 (b) x + 4y – z = 20
3x + 3y + 2z = –3 3x + 2y + z = 8
–2x + 5y + 3z = 3 2x – 3y + 2z = –16

5x – 2y – 4z = 3 …… 1 x + 4y – z = 20 …… 1
3x + 3y + 2z = –3 …… 2 3x + 2y + z = 8 …… 2
–2x + 5y + 3z = 3 …… 3 2x – 3y + 2z = –16 …… 3

2 × 2: 6x + 6y + 4z = –6 1: x + 4y – z = 20 BAB 3
1: 5x – 2y – 4z = 3 2: 3x + 2y + z = 8
Tambah/Add: 4x + 6y = 28 …… 4
Tambah/Add: 11x + 4y = –3 …… 4

2 × 3: 9x + 9y + 6z = –9 2 × 2: 6x + 4y + 2z = 16
3 × 2: –4x + 10y + 6z = 6 3: 2x – 3y + 2z = –16
Tolak/Subtract: 13x – y = –15 …… 5
Tolak/Subtract: 4x + 7y = 32 …… 5

5 × 4: 52x – 4y = –60 4: 4x + 6y = 28
5: 4x + 7y = 32
4: 11x + 4y = –3 Tolak/Subtract: –y = –4
y = 4
Tambah/Add: 63x = –63 Ganti/Replace y = 4 ke dalam/into 4:
4x + 6(4) = 28
x = –1 4x = 4
x = 1
Ganti/Replace x = –1 ke dalam/into 5: Ganti/Replace x = 1, y = 4 ke dalam/into 1:
13(–1) – y = –15 1 + 4(4) – z = 20
y = 2 –z = 3
Ganti/Replace x = –1, y = 2 ke dalam/into 2: z = –3
3(–1) + 3(2) + 2z = –3 Maka/Thus, x = 1, y = 4, z = –3
2z = –6
z = –3
Maka/Thus, x = –1, y = 2, z = –3

(c) 2y – z = 7 (d) 2x + y = 2
x + 2y + z = 17 x+y–z=4
2x – 3y + 2z = –1 3x + 2y + z = 0

2y – z = 7 …… 1 2x + y = 2 …… 1
x + 2y + z = 17 …… 2 x + y – z = 4 …… 2
2x – 3y + 2z = –1 …… 3 3x + 2y + z = 0 …… 3

2 × 2: 2x + 4y + 2z = 34 2: x + y – z = 4
3: 3x + 2y + z = 0
3: 2x – 3y + 2z = –1 Tambah/Add: 4x + 3y = 4 …… 4

Tolak/Subtract: 7y = 35

y = 5 1 × 2: 4x + 2y = 4

Ganti/Replace y = 5 ke dalam/into 1: 4: 4x + 3y = 4
2(5) – z = 7
Tolak/Subtract: –y = 0
z = 3
y = 0

Ganti/Replace y = 5, z = 3 ke dalam/into 2: Ganti/Replace y = 0 ke dalam/into 1: 2x + 0 = 2
x + 2(5) + 3 = 17 x=1

x = 4 Ganti/Replace x = 1, y = 0 ke dalam/into 2:
1 + 0 – z = 4
Maka/Thus, x = 4, y = 5, z = 3 z = –3

Maka/Thus, x = 1, y = 0, z = –3

47

  Matematik Tambahan  Tingkatan 4  Bab 3 Sistem Persamaan

(e) x – y + 3z = 8 (f) 2x – 4y + 3z = 17
3x + y – 2z = –2 x + 2y – z = 0

2x + 4y + z = 0 4x – y – z = 6

x – y + 3z = 8 …… 1 2x – 4y + 3z = 17 …… 1
3x + y – 2z = –2 …… 2 x + 2y – z = 0 …… 2

2x + 4y + z = 0 …… 3 4x – y – z = 6 …… 3

1 : x – y + 3z = 8 2 × 2 : 2x + 4y – 2z = 0

2 : 3x + y – 2z = –2 1 : 2x – 4y + 3z = 17

BAB 3 Tambah/Add : 4x + z = 6 …… 4 Tolak/Subtract : 8y – 5z = –17 …… 4

2 × 4 : 12x + 4y – 8z = –8 2 × 4 : 4x + 8y – 4z = 0

3 : 2x + 4y + z = 0 3 : 4x – y – z = 6

Tolak/Subtract : 10x – 9z = –8 …… 5 Tolak/Subtract : 9y – 3z = –6 …… 5

4 × 9: 36x + 9z = 54 4 × 3: 24y – 15z = –51
5 × 5: 45y – 15z = –30
5 : 10x – 9z = –8 Tolak :
–21y = –21
Tambah/Add: 46x = 46 y = 1

x = 1

Ganti/Replace x = 1 ke dalam/into 4: Ganti/Replace y = 1 ke dalam/into 5:

4(1) + z = 6 9(1) – 3z = –6

z = 2 z = 5

Ganti/Replace x = 1, z = 2 ke dalam/into 1: Ganti/Replace y = 1, z = 5 ke dalam/into 2:

1 – y + 3(2) = 8 x + 2(1) – 5 = 0

y = –1 x = 3

Maka/Thus, x = 1, y = –1, z = 2 Maka/Thus, x = 3, y = 1, z = 5

(g) 2x + y – 2z = –1 (h) x + 3y + 5z = 20
3x – 3y – z = 5 y – 4z = –16

x – 2y + 3z = 6 3x – 2y + 9z = 36

2x + y – 2z = –1 …… 1 x + 3y + 5z = 20 …… 1
3x – 3y – z = 5 …… 2 y – 4z = –16 …… 2

x – 2y + 3z = 6 …… 3 3x – 2y + 9z = 36 …… 3

2 × 2 : 6x – 6y – 2z = 10 1 × 3 : 3x + 9y + 15z = 60

1 : 2x + y – 2z = –1 3 : 3x – 2y + 9z = 36

Tolak/Subtract : 4x – 7y = 11 …… 4 Tolak/Subtract : 11y + 6z = 24 …… 4

2 × 3 : 9x – 9y – 3z = 15 2 × 11: 11y – 44z = –176

3 : x – 2y + 3z = 6 4 : 11y + 6z = 24

Tambah/Add : 10x – 11y = 21 …… 5 Tolak/Subtract : –50z = –200

4 × 10: 40x – 70y = 110 z = 4

5 × 4: 40x – 44y = 84 Ganti/Replace z = 4 ke dalam/into 2:

Tolak/Subtract : –26y = 26 y – 4(4) = –16

y = –1 y = 0

Ganti/Replace y = –1 ke dalam/into 4: Ganti/Replace y = 0, z = 4 ke dalam/into 1:
x + 3(0) + 5(4) = 20
4x – 7(–1) = 11
x = 0
x = 1

Ganti/Replace x = 1, y = –1 ke dalam/into 2: Maka/Thus, x = 0, y = 0, z = 4

3(1) – 3(–1) – z = 5

z = 1

Maka/Thus, x = 1, y = –1, z = 1

48

  Matematik Tambahan  Tingkatan 4  Bab 3 Sistem Persamaan 

3. Selesaikan setiap yang berikut. TP 5

Solve each of the following.

CONTOH

Wahidah, Atikah dan Faruq memesan tiga gabungan makanan yang berlainan di sebuah restoran. Wahidah BAB 3

memesan dua bahagian nasi goreng dan satu bahagian ayam goreng. Atikah memesan satu bahagian nasi

goreng, satu bahagian ayam goreng dan satu bahagian sup cendawan. Faruq memesan satu bahagian sup

cendawan dan dua bahagian ayam goreng. Harga makanan yang dipesan oleh Wahidah, Atikah dan Faruq

masing-masing ialah RM20, RM21 dan RM26. Cari harga bagi satu bahagian ayam goreng.

Wahidah, Atikah and Faruq ordered three combinations of food in a restaurant. Wahidah ordered two parts of fried rice
and one part of fried chicken. Atikah ordered one part of fried rice, one part of fried chicken and one part of mushroom
soup. Faruq ordered one part of mushroom soup and two parts of fried chicken. The cost of the food ordered by Wahidah,
Atikah and Faruq are RM20, RM21 and RM26 respectively. Find the price of one part of fried chicken.

Penyelesaian:

Katakan x = harga satu bahagian nasi goreng 4 : x – z = –1
5 : 4x – z = 14
Let price of one part of fried rice
Tolak: –3x = –15
y = harga satu bahagian ayam goreng Subtraction x = 5

price of one part of fried chicken Gantikan x = 5 ke dalam 4/ Substitute x = 5 into 4:
5 – z = –1
z = harga satu bahagian sup cendawan z = 6
Gantikan x = 5 ke dalam 1/ Substitute x = 5 into 1:
price of one part of mushroom soup 2(5) + y = 20
y = 10
2x + y = 20 …… 1 1 Selesaikan ketiga-tiga Maka, harga bagi satu bahagian ayam goreng ialah
x + y + z = 21 …… 2 persamaan. RM10.
2y + z = 26 …… 3
List all three equations. Thus, the price of one part of fried chicken is RM10.

1 : 2x + y = 20
2 : x + y + z = 21

Tolak/ Subtraction : x – z = –1 …… 4 2 Turunkan
1 × 2 : 4x + 2y = 40 kepada dua
pemboleh ubah.
Reduce to two
3 : 2y + z = 26 variables.

Tolak/ Subtraction : 4x – z = 14 …… 5

Sebanyak 360 kupon telah dijual semasa karnival di sebuah sekolah. Harga kupon ialah RM8, RM10 dan
RM12 dan jumlah pendapatan daripada jualan kupon ialah RM3 500. Gabungan jualan kupon berharga
RM8 dan RM10 adalah lima kali bilangan kupon RM12. Cari bilangan kupon bagi setiap jenis yang dijual.

There were 360 coupons sold during a school carnival. The coupon prices were RM8, RM10 and RM12 and the total income
from the coupon sales was RM3 500. The combined number of RM8 coupons and RM10 coupons sold was five times the
number of RM12 coupons sold. Find the number of coupons of each type sold.

Katakan x = bilangan kupon / number of coupons RM8 Gantikan/Replace z = 60 ke dalam/into 2:
Let say y = bilangan kupon / number of coupons RM10 8x + 10y + 12(60) = 3 500

z = bilangan kupon / number of coupons RM12 8x + 10y = 2 780 …… 5

x + y + z = 360 …… 1 4 × 8 : 8x + 8y = 2 400
8x + 10y + 12z = 3 500 …… 2
x + y = 5z …… 3 5 : 8x + 10y = 2 780

Tolak/Subtract : –2y = –380

1 : x + y + z = 360 y = 190

3 : x + y – 5z = 0 Gantikan/Replace y = 190, z = 60 ke dalam/into 4:
x + 190 = 300
Tolak/Subtract : 6z = 360
x = 110
z = 60

Gantikan/Replace z = 60 ke dalam/into 1:

x + y + 60 = 360 Maka/Thus, x = 110, y = 190, z = 60

x + y = 300 …… 4

49

  Matematik Tambahan  Tingkatan 4  Bab 3 Sistem Persamaan

Persamaan Serentak yang melibatkan Satu Persamaan Linear dan Satu Persamaan

3.2 NOTTAakIMLiBnAeaSrAN
Simultaneous Equations involving One Linear Equation and One Non-Linear Equation
NOTA IMBASAN

BAB 3

4. Selesaikan persamaan serentak berikut menggunakan kaedah penggantian.  TP 4

Solve the following simultaneous equations using the substitution method.

CONTOH 1

3x + y = 5 …… 1
8y2 – 6x = –7 …… 2
Penyelesaian:

Dari 1 / From 1:   y = 5 − 3x …… 3 Jadikan x sebagai perkara rumus.
3 → 2: 8(5 – 3x)2 – 6x + 7 = 0 Make x as the subject of formula..
8(25 – 15x – 15x + 9x2) – 6x + 7 = 0 Gantikan 3 dalam 2.
Substitute 3 into 2.
200 – 240x + 72x2 – 6x + 7 = 0

72x2 − 246x + 207 = 0

24x2 − 82x + 69 = 0 Pemfaktoran.
(2x – 3)(12x – 23) = 0 Factorisation.

x = 3 x = 23 Gantikan nilai x ke dalam 3.
2 12 Substitute the values of x into 3.
3 y = 5 – 3112232
y = 5 – 31 2 2

= 1 = – 3
2 4

\  x = 3 ,  y = 1
dan/ and 2 2

  x = 1223,  y = – 3
4

50

  Matematik Tambahan  Tingkatan 4  Bab 3 Sistem Persamaan 

Selesaikan persamaan serentak berikut menggunakan kaedah penghapusan. TP 4

Solve the following simultaneous equations using the elimination method.

CONTOH 2

4x – y = 6 …… 1
x2 + 3xy = 5 …… 2

Penyelesaian:

1 × 3x: 12x2 – 3xy = 18x …… 3   Hapuskan 3xy
Eliminate 3xy
2 × 3: 13x2 = 5 + 18x

13x2 – 18x – 5 = 0 BAB 3

x = –b ± √b2 – 4ac   Guna rumus kuadratik
2a Use quadratic formula

= –(–18) ± √(–18)2 – 4(13)(–5)
2(13)

= 18 ± √584
26

x = 18 + √584 x = 18 – √584
26 26

= 1.622 = −0.237

y = 4x − 6 y = 4(−0.237) − 6
= 4(1.622) − 6 = −6.948
= 0.488

\  x = 1.62,   y = 0.488
dan/ and

 x = −0.237,   y = −6.948



Selesaikan persamaan serentak berikut menggunakan kaedah penghapusan atau kaedah penggantian. TP 4

Solve the following simultaneous equations using elimination method or substitution method.

(a) x + y – 12 = 0
y2 – 8x = 9

x + y – 12 = 0 …… 1 Gantikan nilai-nilai y ke dalam 3:

y2 – 8x = 9 …… 2 Replace the values of y into 3:

Daripada 1: x = 12 – y …… 3 Apabila/When y = 7
From x = 12 – 7 = 5

Gantikan/Replace 3 ke dalam/into 2: Apabila/When y = –15
x = 12 – (–15) = 27
y2 – 8(12 – y) = 9
Maka, x = 5, y = 7 dan x = 27, y = –15
y2 – 96 + 8y – 9 = 0
Thus and
y2 + 8y – 105 = 0

(y – 7)(y + 15) = 0

y – 7 = 0   atau y + 15 = 0

y = 7   or y = –15

51

  Matematik Tambahan  Tingkatan 4  Bab 3 Sistem Persamaan

(b) x – 2y + 1 = 0
x2 – xy – 3 = 0

x – 2y + 1 = 0 …… 1 Gantikan nilai-nilai y ke dalam 3:

x2 – xy – 3 = 0 …… 2 Replace the values of y into 3:

Daripada/From 1: x = 2y – 1 …… 3 Apabila/When y = – 1
2
Gantikan/Replace 3 ke dalam/into 2:

(2y – 1)2 – (2y – 1)y – 3 = 0 x = 2 1– 1 2 – 1 = –2
2
4y2­ – 4y + 1 – 2y2 + y – 3 = 0

2y2 – 3y – 2 = 0 Apabila/When y = 2
x = 2(2) – 1 = 3
BAB 3 (2y + 1)(y – 2) = 0

2y + 1 = 0   atau/or   y – 2 = 0 Maka/Thus, x = –2, y =– 1 dan/and x = 3, y = 2
2
y = – 1 y = 2
2

(c) 3 + 2y – x = 0
 3x2 + 4y2 = 5 – 3xy

3 + 2y – x = 0 …… 1 Gantikan nilai-nilai y ke dalam 3:

3x2 + 4y2 = 5 – 3xy …… 2 Replace the values of y into 3:

Daripada/From 1: x = 3 + 2y …… 3 Apabila/When y = –0.8083
x = 3 + 2(–0.8083)
Gantikan/Replace 3 ke dalam/into 2: = 1.383
Apabila/When y = –1.2371
3(3 + 2y)2 + 4y2 = 5 – 3(3 + 2y)y x = 3 + 2(–1.2371)
= 0.5258
3(9 + 12y + 4y2) + 4y2 = 5 – 9y – 6y2
Maka/Thus, x = 1.383, y = –0.8083
27 + 36y + 12y2 + 4y2 – 5 + 9y + 6y2 = 0 dan/and x = 0.5258, y = –1.2371

22y2 + 45y + 22 = 0

 y = –45 ± √(45)2 – 4(22)(22)
2(22)

 y = –45 + √89 atau/or  y= –45 – √89
44 44

 y = –0.8083 atau   y = –1.2371


(d) 4x – y – 5 = 0
2x2 – 3y2 + 19 = 0

4x – y – 5 = 0 …… 1 Gantikan nilai-nilai x ke dalam 3:

2x2 – 3y2 + 19 = 0 …… 2 Replace the values of x into 3:

Daripada/From 1: y = 4x – 5 …… 3 Apabila/When x = 14
23
Gantikan/Replace 3 ke dalam/into 2:

2x2 – 3(4x – 5)2 + 19 = 0 y = 4 1 14 2 – 5
23
2x2 – 3(16x2 – 40x + 25) + 19 = 0

2x2 – 48x2 + 120x – 75 + 19 = 0 = – 59
23
–46x2 + 120x – 56 = 0

23x2 – 60x + 28 = 0 Apabila/When x = 2

(23x – 14)(x – 2) = 0 y = 4(2) – 5

23x – 14 = 0 atau/or x – 2 = 2 = 3
14
x = 23 x = 2 Maka/Thus, x = 14 , y = – 59 dan/and x = 2, y = 3
23 23

52

  Matematik Tambahan  Tingkatan 4  Bab 3 Sistem Persamaan 

(e) x – 3y = 5

x2 – xy + y2 – 6 = 0

x – 3y = 5 …… 1 Gantikan nilai-nilai y ke dalam 3:

x2 – xy + y2 – 6 = 0 …… 2 Replace the values of y into 3:

Daripada/From 1: x = 3y + 5 …… 3 Apabila/When y = –1.0969
x = 3(–1.0969) + 5
Gantikan/Replace 3 ke dalam/into 2: = 1.7093
Apabila/When y = –2.4745
(3y + 5)2 – (3y + 5)y + y2 – 6 = 0 x = 3(–2.4745) + 5
= –2.4235
9y2 + 30y + 25 – 3y2 – 5y + y2 – 6 = 0
Maka/Thus, x = 1.7093, y = –1.0969
7y2 + 25y + 19 = 0 dan/and x = –2.4235, y = –2.4745

 y = –25 ± √(25)2 – 4(7)(19) BAB 3
2(7)

 y = –25 + √93 atau/or  y = –25 – √93
14 14

 y = –1.0969 atau   y = –2.4745

(f) 3x + 2y = 10

 3x + 2 =5
y

3x + 2y = 10 …… 1 Gantikan nilai-nilai y ke dalam 3:
3y + 2x = 5xy …… 2 Replace the values of y into 3:

Daripada/From 1: y = 10 – 3x …… 3 Apabila/When x = 2
2 3
3 1 2 2
Gantikan/Replace 3 ke dalam/into 2: 10 – 3

31 10 – 3x 2 + 2x = 5x1 10 – 3x 2 y= 2 =4
2 2
Apabila/When x = 3
30 – 9x + 4x = 50x – 15x2 10 – 3(3) 1
15x2 – 55x + 30 = 0 y= 2 = 2
3x2 – 11x + 6 = 0
(3x – 2)(x – 3) = 0 Maka/Thus, x = 2 , y = 4 dan/and x = 3, y = 1
3 2
2
  x = 3   atau/or  x = 3

(g)    x + 3y = 1

4y + 3x = –13
x y

x + 3y = 1 …… 1 Gantikan nilai-nilai y ke dalam 3:
4y2 + 3x2 = –13xy …… 2
Replace the values of y into 3:

Daripada/From 1: x = 1 – 3y …… 3 Apabila/When y = 3
8

Gantikan/Replace 3 ke dalam/into 2: 1 2 x = 1 – 3  3
– 1 8
4y2 + 3(1 – 3y)2 = –13(1 – 3y)y = 8

4y2 + 3(1 – 6y + 9y2) = –13y + 39y2

4y2 + 3 – 18y + 27y2 + 13y – 39y2 = 0 Apabila/When y = –1

–8y2 – 5y + 3 = 0 x = 1 – 3(–1)

8y2 + 5y – 3 = 0 = 4

(8y – 3)(y + 1) = 0 1 3
8 8
8y – 3 = 0   atau/or y + 1 = 0 Maka/Thus, x =– , y = dan/and x = 4, y = –1

y = 3 y = –1
8

53

  Matematik Tambahan  Tingkatan 4  Bab 3 Sistem Persamaan

Selesaikan persamaan serentak berikut menggunakan kaedah perwakilan graf.   TP 5

Solve the following simultaneous equations using the graphical representation method.

CONTOH 3

y + 2x = 7
y2 = 4x + 1
Penyelesaian:
Bina satu jadual nilai / Construct a table of values

BAB 3 x 012345 6 7
–5 –7
y = 7 – 2x 7 5 3 1 –1 –3 ±5 ±5.39

y = ±√4x + 1 ±1 ±2.24 ±3 ±3.61 ±4.12 ±4.58

Bina satu graf / Construct a graph

y

8 x
7
6
5
4
3 (2, 3)
2
1

–1 0 1 2 3 4 5 6 7 8
–1
–2
–3
–4
–5 (6, –5)
–6
–7

Titik persilangan mewakili penyelesaian persamaan serentak:

Point of intersection represents the solution of simultaneous equations:

(2,3) dan /and (6,−5).

54

  Matematik Tambahan  Tingkatan 4  Bab 3 Sistem Persamaan 

Selesaikan persamaan serentak berikut menggunakan kaedah perwakilan graf.  5

Solve the following simultaneous equations using the graphical representation method.

(h) x 0 1234 567
7 5 3 1 –1 –3 –5 –7
y = 7 – 2x
– 9 3 0.33 –1.5 –3 –4.33 –5.57
y= 10 – x2
x

y + 2x = 7 y
x2 + xy = 10
0 < x < 7 8 BAB 3
7
6 (2, 3)
5 x
4
3 12345678
2 (5, –3)
1

–1 0
–1
–2
–3
–4
–5
–6
–7

Penyelesaian / Solutions: (2,3),(5,−3)

(i) x –3 –2 –1 0 1 2 3

y=2–x 5 4 3 2 1 0 –1

y = ±√9– x2­ 0 ±2.24 ±2.83 ±3 ±2.83 ±2.24 0

x + y = 2 (–0.871, 2.871) 3
x2 + y2 = 9 2
–3 < x < 3

1

–3 –2 –1 0 12 3
–1
–2 (2.871, –0.871)

–3

Penyelesaian / Solutions: (−0.8,2.85),(2.8,−0.80)
55

  Matematik Tambahan  Tingkatan 4  Bab 3 Sistem Persamaan

5. Selesaikan setiap yang berikut: TP 5

Solve each of the following:

(a) Cari koordinat titik-titik persilangan antara garis lurus y = –3x – 4 dan lengkung xy + 40 = y2.

Find the coordinates of the points of intersection between the line y = –3x – 4 and the curve xy + 40 = y2.

y = –3x – 4 …… 1
xy + 40 = y2 …… 2

Gantikan 1 ke dalam 2: Gantikan nilai-nilai x ke dalam 1:

Replace 1 into 2: Replace the values of x into 1:

BAB 3 x(–3x – 4) + 40 = (–3x – 4)2 Apabila/When x = 2 ,
–3x2 – 4x + 40 = 9x2 + 24x + 16 3
12x2 + 28x – 24 = 0
3x2 + 7x – 6 = 0 y = –31 2 2 – 4
(3x – 2)(x + 3) = 0 3

= – 6

3x – 2 = 0 atau   x + 3 = 0 Apabila/When x = –3,
2 y = –3(–3) – 4
x = 3    or x = –3 = 5

1 2Maka, titik-titik persilangan ialah 2 , –6 dan (–3, 5).
3
1 2Thus, the points of intersections are2
3 , –6 and (–3, 5).

(b) Diberi panjang hipotenus sebuah segi tiga tepat ialah 35 cm dan perimeter segi tiga tepat itu ialah
84 cm. Cari panjang yang mungkin bagi dua sisi yang lain bagi segi tiga tepat itu.

Given the length of the hypotenuse of a right-angled triangle is 35 cm and the perimeter of the right- angled triangle is
84 cm. Find the possible length of the other two sides of the triangle.

x + y + 35 = 84
x + y = 49 …… 1

x2 + y2 = 352
x2 + y2 = 1 225 …… 2

Daripada/From 1: y = 49 – x …… 3 Gantikan nilai-nilai x ke dalam 3:

Gantikan 3 ke dalam 2: Replace the values of x into 3:

Replace 3 into 2: Apabila/When x = 28,
y = 49 – 28
x2 + (49 – x)2 = 1 225 = 21
Apabila/When x = 21,
x2 + (2 401 – 98x + x2) = 1 225 y = 49 – 21
= 28
2x2 – 98x + 1 176 = 0
Maka, panjang sisi-sisi segi tiga ialah 21 cm dan
x2 – 49x + 588 = 0 28 cm.

(x – 28)(x – 21) = 0 Thus, the length of the sides of the triangle are 21 cm and
28 cm.
x – 28 = 0   atau x – 21 = 0

x = 28 x = 21

56

  Matematik Tambahan  Tingkatan 4  Bab 3 Sistem Persamaan 

PRAKTIS SPM 3

Kertas 2 Diberi bahawa SR = x m dan PT = y m. Luas laman

1. Selesaikan persamaan serentak berikut. berbentuk segi empat tepat PQRS ialah 77  m2 dan

SPM Solve the following simultaneous equations. perimeter kawasan berumput ialah 33 m. Kolam

2 01 4 5 + y – 3x = 0,    x2 + y2 – 2y – 25 = 0 renang dengan kedalaman seragam mengandungi
i=tu2.72 ,
Beri jawapan anda betul kepada dua tempat 46.2 m3 air. Dengan menggunakan π cari
kedalaman, dalam m, air dalam kolam
perpuluhan.
It is given that SR = x m and PT = y m. The area of a
Give your answers correct to two decimal places. rectangular backyard is 77 m2 and the perimeter of the

5 + y – 3x = 0 ………… 1 grassy area is 33 m. The swimming pool with uniform

x2 + y2 – 2y – 25 = 0 ………… 2 depth contains 46.2 m3 of water. By using π = 22 , find
7
Daripada/From 1, y = 3x – 5 …… 3 the depth, in m, of water in the pool. BAB 3

Gantikan 3 ke dalam 2.
Replace 3 into 2:
L uas laman/backyard area = 77
x2 + (3x – 5)2 – 2(3x – 5) – 25 = 0 (y + x)x = 77
xy + x2 = 77 …… 1
x2 + 9x2 – 30x + 25 – 6x + 10 – 25 = 0

10x2 – 36x + 10 = 0 Perimeter kawasan berumput = 33

5x2 – 18x + 5 = 0 Perimeter of grassy area = 33

x = –(–18) ± (–18)2­ – 4(5)(5) y + (y + x) + x + 1 πj = 33
2(5) 2

= 0.30 atau 3.30 1 2 2y + 2x + 1   22 x = 33
2 7
Gantikan nilai-nilai x ke dalam 3.
2y + 2x + 171x = 33
Replace the values of x into 3.
14y + 14x + 11x = 231
Apabila/When  x = 0.30,  
y = 3(0.30) – 5 14y + 25x = 231 …… 2
= –4.10

Apabila/When  x = 3.30, Daripada/From 2:  y = 231 – 25x …… 3
y = 3(3.30) – 5 14
= 4.90
Gantikan 3 ke dalam 1:
Maka, x = 0.30, y = –4.10 dan x = 3.30, y = 4.90.
Replace 3 into 1:
Thus, x = 0.30, y = –4.10 and x = 3.30, y = 4.90.
231 – 25x
14
1 2 x + x2 = 77

2. Rajah menunjukkan pelan bagi laman belakang 231x – 25x2 + 14x2 = 1 078

SPM sebuah rumah banglo berbentuk segi empat tepat –11x2 + 231x – 1 078 = 0
2018 PQRS. Laman itu terdiri daripada sebuah kolam
x2 – 21x + 98 = 0
renang berbentuk sukuan bulatan dan kawasan
(x – 14)(x – 7) = 0
berumput PQRUT.
x – 14 = 0   atau x – 7 = 0
Diagram shows the plan of a rectangular backyard of a x = 14  or x=7
bungalow PQRS. The backyard consists of a swimming
pool in a shape of a quadrant of a circle and a grassy Gantikan nilai-nilai x ke dalam 3:
area PQRUT.
Replace the values of x into 3.

PT Apabila/When x = 14,
S 231 – 25(14) 17
y= 14 =– 2 (tidak diterima)

Apabila/When x = 7,

U y= 231 – 25(7) =4
Q 14

R Maka/Thus, x = 7, y = 4

57

  Matematik Tambahan  Tingkatan 4  Bab 3 Sistem Persamaan

Katakan kedalaman air = d 4. Selesaikan persamaan serentak berikut.
Let the depth of water = d
SPM Solve the following simultaneous equations.
Isi padu/Volume = luas/area × d
2015 3 + y – 2x = 0, 3x2 + 2y2 – 4xy = 9

46.2 = 1 pj2d 3x2 + 2y2 – 4xy = 9 …… 1
4 3 + y – 2x = 0

1 2 1 22 (7)2d ⇒ y = 2x – 3 …… 2
46.2 = 4   7

46.2 = 154  d Gantikan/Replace 2 ke/into 1.
4 3x2 + 2(2x − 3)2 – 4x(2x − 3) = 9
3x2 + 2(4x2 – 12x + 9) – 8x2 + 12x – 9 = 0
d = 1.2 3x2 + 8x2 – 24x + 18 – 8x2 + 12x – 9 = 0 ÷3
Maka/Thus, d = 1.2 meter 3x2 – 12x + 9 = 0
x2 – 4x + 3 = 0
BAB 3 (x – 1)(x – 3) = 0

x – 1 = 0 , x – 3 = 0
x = 1 , x = 3

3. Diberi persamaan berikut: Apabila/When x = 1,  y = 2(1) – 3
= –1
SPM Given the following equations: Apabila/When x = 3,  y = 2(3) – 3
=3
2017 P = x – 2y Maka/Thus, x = 1, y = 1 dan/and x = 3, y = 3.

BUKAN Q = 3x – y + 1 5. Selesaikan persamaan serentak x + 3y = 2 dan
R = x2 + 4y2 4 3
RUTIN SPM x – y = 8. Beri jawapan anda betul kepada tiga
2019


Cari nilai-nilai x dan y jika Q = R = 2P. tempat perpuluhan.

Find the values of x and y if Q = R = 2P.

Q = R = 2P Solve the simultaneous equations x + 3y = 2 and
4 3
3x – y + 1 = x2 + 4y2 = 2(x – 2y) x – y = 8. Give you answers correct to three decimal

3x – y + 1 = x2 + 4y2 …… 1 places.

3x – y + 1 = 2(x – 2y) …… 2 x + 3y = 2

Daripada 2, 3x – y + 1 = 2x – 4y x = 2 – 3y …… 1

From 3x – 2x = y – 4y – 1 4 – 3 = 8
x y
x = –3y – 1 …… 3

Gantikan 3 ke dalam 1. × xy → 4y – 3x = 8xy …… 2

Replace 3 into 1: 1 → 2: 4y – 3(2 – 3y) = 8y(2 – 3y)

3(–3y – 1) – y + 1 = (–3y – 1)2 + 4y2 4y – 6 + 9y = 16y – 24y2

–9y – 3 – y + 1 = 9y2 + 6y + 1 + 4y2 24y2 – 3y – 6 = 0
y = –b
13y2 + 16y + 3 = 0 ± b­2 – 4ac
2a
(13y + 3)(y + 1) = 0

y = – 3   atau   y = –1 = –(–1) ± (–1)2­ – 4(8)(–2)
13 2(8)

Gantikan nilai-nilai y ke dalam 3. = 1 ± 65
Replace the values of y into 3: 16
3
1 2Apabila/When y = – 13 , Apabila/When y = –1, 1 + 65 1 – 65
x = –3(–1) – 1 16 16
x = –3 – 3 –1 =2 y = y =
13
4 = 0.566 = –0.441
=– 13
x = 2 – 3(0.566) x = 2 – 3(–0.441)

Maka/Thus, x = – 4 , y = – 3 dan/and x = 2, y = –1. = 0.302 = 3.323
13 13
\  x = 0.302,  y = 0.566 Praktis
dan/and x = 3.323,  y = –0.441 SPM
Ekstra

58

  Matematik Tambahan  Tingkatan 4  Bab 3 Sistem Persamaan 

Sudut KBAT KBAT

Ekstra

1. Diberi (2m, 3n) adalah penyelesaian kepada 2. Suatu jenama minuman tin tertentu dikeluarkan
4 3 13
persamaan serentak y + 2x = –4 dan y – 2x =– 6 , secara pek dengan 6, 12 dan 24 tin setiap jenis
cari nilai-nilai m dan n.
dan masing-masing berharga RM10, RM18 dan

Given that (2m, 3n) is the solution to the simultaneous RM36 setiap pek. Sebuah stor telah menjual
4 3 13
equations y + 2x = –4 and y – 2x =– 6 , find the 14 pek dengan jumlah 162 tin dan menerima
values of m and of n.
bayaran RM248. Cari bilangan pek setiap jenis
x = 2m, y = 3n
yang telah dijual.
y + 2x = –4
A certain brand of canned drinks comes in packages
3n + 4m = –4 of 6, 12 and 24 cans costing RM10, RM18 and RM36
per package respectively. A store sold 14 packages
3n = –4 – 4m containing a total of 162 cans and received RM248. BAB 3
Find the number of packages of each type sold.

n = –4 – 4m …  Katakan x = bilangan pek 6 tin
3 Let y = bilangan pek 12 tin
z = bilangan pek 24 tin
4 – 23x = – 13
y 6

8x – 3y = – 13 x + y + z = 14 …… 1
2xy 6

6(8x – 3y) = –13(2xy) 10x + 18y + 36z = 248 …… 2
48x – 18y = –26xy
48(2m) – 18(3n) = –26(2m)(3n) 6x + 12y + 24z = 162 …… 3
96m – 54n + 156mn = 0 …
3 ÷ 6: x + 2y + 4z = 27

1: x + y + z = 14

Gantikan  kepada : Tolak/Subtract: y + 3z = 13 …… 4

Replace  into : 1 × 5: 5x + 5y + 5z = 70

96m – 54 –4 – 4m  + 156m  –4 – 4m  = 0 2 ÷ 2: 5x + 9y + 18z = 124
3 3
96m – 18(–4 – 4m) + 52m (–4 – 4m) = 0 Tolak/Subtract: –4y – 13z = –54 …… 5

96m + 72 + 72m – 208m – 208m2 = 0 4 × 4: 4y + 12x = 52

208m2 + 40m – 72 = 0 5: –4y – 13z = –54

52m2 + 10m – 18 = 0 Tambah/Add: –z = –2

26m2 + 5m – 9 = 0 z = 2

(13m + 9)(2m – 1) = 0 Ganti z = 2 ke dalam 4:

13m + 9 = 0 , 2m – 1 = 0 Replace z = 2 into 4:

m = – 193 m = 1 y + 3(2) = 13
2
y = 7

n = 1 –4 – 4– 193 n = 1 –4 – 4 1  Ganti y = 7, z = 2 ke dalam 1:
3 3 2
Replace y = 7, z = 2 into 1:

= – 1396 = –2 x + 7 + 2 = 14

x = 5

Maka/Thus, x = 5, y = 7, z = 2

Kuiz 3

59

BAB Indeks, Surd dan Logaritma

4 Indices, Surds and Logarithms

4.1 Hukum Indeks
4.1
Laws of Indices

NOTA IMBASAN

Hukum indeks
Laws of indices

1. Permudahkan setiap ungkapan algebra yang berikut.  TP 1

Simplify each of the following algebraic expressions.

CONTOH (a) m3 × (n–2)4 × (5m)4
(3m2n)3
52n + 5 × 53 – n × 25n + 5
25(2n – 1) = m3 × n–8 × 625m4
27m6n3
52n + 5 × 53 – n × 52(n + 5)
= 52(2n – 1) = 625m7n–8
27m6n3
2n + 5 + 3 – n + 2n + 10
= 625  m7 – 6n–8 – 3
= 5 54n – 2 27
= 53n + 18 − (4n − 2)
625
= 52 – n = 27  mn−11

60

  Matematik Tambahan  Tingkatan 4  Bab 4 Indeks, Surd dan Logaritma 

(b) x2m + 2 × (yn − 1)3 ÷ (xyn)4 (c) 31 3x2 24 × 1 16y2 2342
= x2m + 2 × y3n − 3 ÷ x4 y4n 4y3 9x3
= x2m + 2 − 4 y3n − 3 − 4n
= x2m − 2 y −3 − n 81x8 4096y6 2
256y12 729x9
1 2 = ×

= 1 16 22
9xy6

= 256
81x2 y12

2. Permudahkan setiap ungkapan algebra yang berikut.  TP 2

Simplify each of the following algebraic expressions.

CONTOH (a) 8 2 x + 43 + x −16 1 x + 1
3 2

7(2n) − 8 1 n − 1 + 4 1 n + 3 = (23) 2 x + (22)3 + x −(24) 1 x + 1 BAB 4
3 2 3 2

= 7(2n)−(23) 1 n − 1 + (22) 1 n + 3 = 22x + 26 + 2x − 22x + 4
3 2

= 7(2n)−2n − 3 + 2n + 6 = 22x + 26(22x) − 24(22x)

= 7(2n)−2n(2−3) + 2n(26) = 22x(1 + 26 − 24)

= 2n17 – 1 + 642 = 4x(1 + 64 − 16)
8
7 = 49(4x)
8
= 70 (2n)

= 567 (2n)
8

= 567(2n – 3)

(b) a3 + n × (a2)n − 1 (c) 5n + 3 +125 1 (n – 1) − 25 1 n
a4 3 2
−n

a3 + n × a2n − 2 = 5n(53) + (53) 1 (n – 1) – (52) 1 n
a4 − n 3 2
=
= 5n(53) + 5n − 1 − 5n

= a3 + n + 2n − 2 − (4 − n) 1= 5n 53 + 1 – 12
5
= a3n + 1 − 4 + n 1
5
= a4n − 3 = 124 (5n)

= 621 (5n)
5

= 621(5n − 1)

3. Buktikan.   TP 3

Prove.

CONTOH (a) 34 – 2x = 81
9x
8x
23x – 4 = 16 34 − 2x = 34 ÷ 32x

23x – 4 = 23x ÷ 24 = 34 = 81
32x 9x
23x 8x
= 24 = 16

61

  Matematik Tambahan  Tingkatan 4  Bab 4 Indeks, Surd dan Logaritma

(b) 7n – 2m = 7n (c) 43m – 2 = 64m
49m 16

7n − 2m = 7n ÷ 72m 43m − 2 = 43m ÷ 42

= 7n = 7n = 43m = 64m
72m 49m 42 16

4. Tunjukkan.   TP 3

Show.

BAB 4 CONTOH (a) Tunjukkan bahawa 9n + 9n + 3 − 9n + 2 boleh dibahagi
tepat oleh 11 bagi semua integer positif n.
Tunjukkan bahawa 7n + 7n + 2 + 7n + 3 boleh dibahagi
tepat oleh 3 bagi semua integer positif n. Show that 9n + 9n + 3 − 9n + 2 is divisble by 11 for all
positive integers n.
Show that 7n + 7n + 2 + 7n + 3 is divisible by 3 for all positive
integers n. 9n + 9n + 3 − 9n + 2
= 9n(1) + 9n(93) − 9n(92)
7n + 7n + 2 + 7n + 3 = 9n(1 + 729 − 81)
=7n(1) + 7n­(72) + 7n(73) = 9n(649)
=7n(1 + 49 + 343) = 9n(11 × 59)
=7n(393)
=7n(3 × 131) Boleh dibahagi tepat oleh 11 kerana 11 ialah satu
faktor.
Boleh dibahagi tepat oleh 3 kerana 3 ialah satu
faktor. Divisible by 11 since 11 is a factor.

Divisble by 3 since 3 is a factor.

(b) Tunjukkan bahawa 5n + 2 – 5n + 1 + 5n + 3 boleh (c) Tunjukkan bahawa 6n + 3 + 6n + 2 − 6n + 1 boleh

dibahagi tepat oleh 29 bagi semua integer positif dibahagi tepat oleh 41 bagi semua integer positif

n. n.

Show that 5n + 2 – 5n + 1 + 5n + 3 is divisble by 29 for all Show that 6n + 3 + 6n + 2 − 6n + 1 is divisble by 41 for all
positive integers n. positive integers n.

5n + 2 – 5n + 1 + 5n + 3 6n + 3 + 6n + 2 − 6n + 1
= 5n(52) – 5n(51) + 5n(53) = 6n(63) + 6n(62) − 6n(61)
= 5n(25 – 5 + 125) = 6n(216 + 36 − 6)
= 5n(145) = 6n(246)
= 5n(29 × 5) = 6n(41 × 6)

Boleh dibahagi tepat oleh 29 kerana 29 ialah satu Boleh dibahagi tepat oleh 41 kerana 41 ialah satu

faktor. faktor.
Divisible by 29 since 29 is a factor. Divisible by 41 since 41 is a factor.

62

  Matematik Tambahan  Tingkatan 4  Bab 4 Indeks, Surd dan Logaritma 

5. Selesaikan setiap persamaan yang berikut.  TP 4

Solve each of the following equations.

CONTOH (a) 4x × 22x − 1 = 128

3x + 2 – 3x – 1 = 26 22x × 22x − 1 = 27
27
22x + 2x − 1 = 27

3x(32) – 3x(3–1) = 26 2x + 2x − 1 = 7
27
4x = 8

3x19 1 2 = 26 x = 2
3 27
–

3x1 26 2 = 26
3 27

1 2 26(3x – 1) = 26 1
27

3x – 1 ­ = 3–3 BAB 4
x – 1 = –3
x = –2

(b) 53(x – 1) = 25x – 1 (c) 4x + 2 = 64x + 1
16x

53x + 3 = 52(x – 1) 4x + 2 = (43)x + 1
3x + 3 = 2x – 2 16x

x = –5 4x + 2 – 2x = 43x + 3

2 − x = 3x + 3

4x = –1

x = – 1
4

(d) 8x + 2 = 1 (e) 52x + 1 + 52x − 1 = 26
512x + 1

8x + 2 × 512x + 1 = 1 52x(51) + 52x(5−1) = 26

8x + 2 × 83(x + 1) = 80 152x 5 + 1 2 = 26
5
x + 2 + 3x + 3 = 0
152x 26 2
4x = –5 5 = 26

x = – 5 52x = 5
4
2x = 1

x = 1
2

(f) 62x + 1 + 62x − 1 = 1332 (g) 4x + 2 − 4x − 1 = 252

62x(61)+62x(6−1) = 1332 4x(42) – 4x(4−1) = 252

1 62x 6 + 1 2 = 1332 4x116 – 1 2 = 252
6 4

1 262x37 = 1332 6 4x1 63 2 = 252
6 37 4
62x = 1332 × 4
216 4x = 252 × 63
62x =

62x = 63 4x = 16

2x = 3 4x = 42
3
x = 2 x = 2

63

  Matematik Tambahan  Tingkatan 4  Bab 4 Indeks, Surd dan Logaritma

6. Selesaikan masalah yang berikut.  TP 4

Solve the following problems.

(a) Suhu, T, bagi sebuah aluminium, dalam °C, (b) (b) Nilai sebuah kereta menyusut sebanyak

selepas dipanaskan selama t minit diberikan oleh 8% daripada nilainya pada awal setiap tahun.

T = 28.5(1.1)t. Cari suhunya selepas setengah Jika harga asal kereta itu ialah RM145 0 00,

jam. nilainya p, dalam RM, selepas t tahun diberi oleh
p = 145  000(0.92)t. Carikan harga kereta itu
The temperature, T, of a piece of aluminium, in °C, after selepas 7 tahun.
its heated for t minutes is given by T = 28.5(1.1)t. Find
its temperature after heated for half an hour. The value of a car decreases by 8% from its value at
the beginning of the year. If its original price was
T = 28.5(1.1)t RM145  000, its value p, in RM, after t years is given by p
= 145 000(0.92)t. Find the price of the car after 7 years.
t = 30 minit / minutes

T = 28.5(1.1)30 p = 145000(0.92)t

= 497.31°C t = 7

BAB 4 p = 145000(0.92)7

= 80 887.76

4.2 Hukum Surd

Laws of Surds

NOTA IMBASAN Surd
NOTA IMBASAN Surd

64

  Matematik Tambahan  Tingkatan 4  Bab 4 Indeks, Surd dan Logaritma 

7. Tukar perpuluhan berulang berikut kepada pecahan.   TP 3

Convert the following recurring decimals to fractions.

CONTOH (a) 0.7777…

0.353535…

Penyelesaian: Biar / Let N = 0.7777 …… 1

Biar / Let N = 0.353535 …… 1 10N = 7.7777 …… 2

100N = 35.3535 …… 2 2 – 1: 9N = 7

2 – 1: 99N = 35 N = 7
9
35
N = 99

(b) 0.818181… (c) 0.717171…

Biar / Let N = 0.818181 …… 1 Biar / Let N = 0.717717717 …… 1 BAB 4

100N = 81.8181 …… 2 1000N = 717.717717 …… 2

2 – 1: 99N = 81 2 – 1: 999N = 717

N = 81 = 9 N = 717 = 239
99 11 999 333

8. Tentukan sama ada setiap yang berikut ialah surd. Nyatakan sebab anda.   TP 3

Determine whether each of the foillowing is a surd. State your reason.

CONTOH (a) 5√243 =

4√80 = 2.990697562… 5√243 = 3

Nilainya ialah satu integer. 5√243 bukan sebuah
(Gunakan kalkulator / Use calculator) surd.
The value is an integer. 5√243 is not a surd.
4√80 ialah sebuah surd kerana bukan perpuluhan
berulang.

4√80 is a surd because it is a non-recurring decimal.

(b) 4 16 = (c) 6√88 =
625

4 16 = 2 6√88 = 2.109018772…
625 5
6√88 ialah sebuah surd kerana bukan perpuluhan
Nilainya ialah satu pecahan. 4 16 bukan berulang.
sebuah surd. 625
6√88 is a surd because it is a non-recurrung decimal.

The value is a fraction. 4 16 is not a surd.
625

65

  Matematik Tambahan  Tingkatan 4  Bab 4 Indeks, Surd dan Logaritma

9. Tulis setiap yang berikut sebagai satu surd tunggal.   TP 3

Write each of the following as a single surd.

CONTOH (a) √3  × √11  (b) √78
= √3 × 11  √6
√5x  × √7x  = √33 
= √5x × 7x = 78
= √35x2 6
= x √35 
= √13 

(c) √13m  × √2m  (d) √20n (e) √6 × √10
= √13m × 2m  √32n √5
= √26m2 
= m√26  = 20n = 6 × 10
32n 5
BAB 4
= 5 = √12 
8

10. Permudahkan setiap yang berikut: TP 3

Simplify each of the following:

CONTOH (a) √32  (b) √45 

√12  √32  = √16 × 2  √45  = √9 × 5 
= √16  × √2  = √9  × √5 
Penyelesaian: = 4√2  = 3√5 
√12  = √4 × 3 
= √4  × √3 
= 2√3 

(c) √125  (d) 3√24  (e) 3√135 
3√24  = 3√8 × 3 
√125  = √25 × 5  = 3√8 × 3√3 3√135  = 3√27 × 5 
= √25  × √5  = 23√3 = 3√27 × 3√5
= 5√5  = 33√5


11. Permudahkan setiap yang berikut:   TP 3

Simplify each of the following:

CONTOH 1 (a) 3√2  + 5√2  – √2  (b) 4√3  – 7√3 

3√8 + 2√8 – 4√8 3√2  + 5√2  – √2  4√3  – 7√3 
= (3 + 5 – 1)√2  = (4 – 7)√3 
Penyelesaian: = 7√2  = –3√3 

3√8 + 2√8 – 4√8
= (3 + 2 – 4)√8 
= √8 

66

  Matematik Tambahan  Tingkatan 4  Bab 4 Indeks, Surd dan Logaritma 

CONTOH 2 (c) 13√28  – 7√7  (d) 2√12  + 5√48  – 7√3  

√18  – √8  = 13√4 × 7  – 7√7  = 2√4 × 3  + 5√16 × 3  – 7√3 
= 13√4 √7  – 7√7  = 2√4 √3  + 5√16 √3  – 7√3 
Penyelesaian: = 26√7  – 7√7  = 4√3  + 20√3  – 7√3 
= (26 – 7)√7  = (4 + 20 – 7)√3 
= √9 × 2  – √4 × 2  Tukar ke sebutan = 19√7 = 17√3
surd yang serupa.
= 3√2  – 2√2

= (3 – 2)√2  Change to equal
= √2  surd term.

(e) 3√20  – √5  – 2√45  (f) 2√6  + √150  – 3√54  (g) √2  × √3  + 4√6  BAB 4

3√20  – √5  – 2√45  2√6  + √150  – 3√54  √2  × √3  + 4√6 
= 3√4 × 5  – √5  – 2√9 × 5  = 2√6  + √25 × 6  – 3√9 × 6  = √2 × 3  + 4√6 
= 3√4 √5 – √5  – 2√9 √5  = 2√6  + 5√6  – 9√6  = √6  + 4√6 
= 6√5  – √5  – 6√5  = (2 + 5 – 9)√6  = (1 + 4)√6 
= (6 – 1 – 6)√5  = –2√6  = 5√6 
= –√5 

CONTOH 3 (h) 8 (i) 20
25 9
28
9 8 = √8  20 = √20 
25 5 9 3
Penyelesaian:
= √4 × 2  = √4 × 5 
28 = √28  5 3
9 3
= √4  × √2  = √4  × √5 
= √4 × 7  5 3
3
= 2√2  = 2√5 
= √4  × √7  5 3
3

= 2√7 
3

(j) 18 (k) 3 24 (l) 3 8
75 27 16

18 = 6 3 24 = 3√24  3 8 = 3 1
75 25 27 3√27  16 2

= √6  = 3√8 × 3  = 1 × 3√4 
√25  3 3√2 3√4 

= √6  = 3√8  × 3√3 = 3√4 
5 3 3√8 

= 2 3√3  = 3√4 
3 2

67

  Matematik Tambahan  Tingkatan 4  Bab 4 Indeks, Surd dan Logaritma

12. Permudahkan setiap yang berikut: TP 3

Simplify each of the following: (a) 5 (b) 2
√2 3
CONTOH 1

2 5 = 5 √2 2 √2 √3
√2 √2 √2 3 √3 √3
√3 × = ×

Penyelesaian: 5√2
2 √6
2 = 2 × √3 = = 3
√3 √3 √3

= 2√3
3

BAB 4 (c) √5  (d) √7  (e) 6
√8  √12  √5

√5  = √5  √7  = √7  6 = 6 × √5
√8  √4 × 2  √12  √4 × 3  √5 √5 √5

= √5  × √2  = √7  × √3  = 6√5
2√2  √2  2√3  √3  5

= √10 = √21
4 6

CONTOH 2 (f) 3√2 (g) 6√7
5√3 5√27
3√5
3√2 = 3√2 × √3 6√7 = 6√7 
4√3 5√3 5√3 √3 5√27 5√9 × 3
Penyelesaian:
= 3√6 = 6√7 
3√5 = 3√5 × √3  15 5√9 √3 
4√3 4√3 √3 
= √6 = 6√7 × √3
= 3√15 5 15√3 √3
12
= 6√21
= √15 45
4
= 2√21
15

(h) 5√5 (i) 8√5 (j) 4√45
3√2 3√12 –6√20

5√5 = 5√5 × √2 8√5 = 8√5 3  4√45 = 4√9 × 5 
3√2 3√2 √2 3√12 3√4 × –6√20 –6√4 × 5

= 5√10 = 8√5 = 4√9 √5 
6 6√3 –6√4 √5 

= 8√5 × √3 = 12√5
6√3 √3 –12√5

= 8√15 = –1
18

= 4√15
9

68

  Matematik Tambahan  Tingkatan 4  Bab 4 Indeks, Surd dan Logaritma 

13. Permudahkan setiap yang berikut: TP 3

Simplify each of the following:

CONTOH 1 CONTOH 2

3 4
√2 + √5 √3 – √2

Penyelesaian: Penyelesaian:

3 (√2 – √5) Darabkan dengan konjugat. 4 (√3 + √2)
Multiply with conjugate.
(√2 + √5) (√2 – √5) (√3 – √2) (√3 + √2)

= 3√2 – 3√5 = 4√3 + 4√2
2–5 3–2

= 3√2 – 3√5 = 4√3 + 4√2
–3

= √5 – √2 BAB 4

(a) 3 (b) √3
√7 – √2 √8 + √5

= 3 (√7 + √2) = √3 (√8 – √5)
(√7 – √2) (√7 + √2) (√8 + √5) (√8 – √5)

= 3√7 + 3√2 = √3 √8 – √3 √5
7–2 8–5

= 3√7 + 3√2 = √3 √4 × 2 – √15
5 3

= √3 √4 √2 – √15
3

= 2√6 – √15
3

(c) 4 (d) 3
√7 – 1 2√3 + 3√5

= 4 (√7 + 1) = 3 (2√3 – 3√5)
(√7 – 1) (√7 + 1) (2√3 + 3√5) (2√3 – 3√5)

= 4√7 + 4 = 3(2√3 – 3√5)
7–1 4(3) – 9(5)

= 4√7 + 4 = 3(2√3 – 3√5)
6 –33

= 2(√7 + 1) = 2√3 – 3√5
3 –11

= 3√5 – 2√3
11

69

  Matematik Tambahan  Tingkatan 4  Bab 4 Indeks, Surd dan Logaritma

(e) 3√2 (f) 2√20
2√6 – √10 1 – √5

= 3√2 (2√6 + √10) = 2√20 (1 + √5)
(1 – √5) (1 + √5)
(2√6 – √10) (2√6 + √10)

= 3√2(2√6 + √10) = 2√20 + 2√100
4(6) – 10 1–5

= 6√12 + 3√20 = 2√4 × 5+ 2(10)
14 –4

= 6√4 × 3 + 3√4 × 5 = 4√5 + 20
14 –4

= 12√3 + 6√5 = –√5 – 5
14

BAB 4 = 6√3 + 3√5
7

(g) √y  – 4 (h) 8√m  + 5√n 
√y  + 2 √m  – √n 

= √y  – 4 (√y  – 2) = (8√m  + 5√n) (√m  + √n )
(√y  + 2) (√y  – 2) (√m  – √n ) (√m  + √n )

= y – 2√y  – 4√y  + 8 = 8m + 8√mn  + 5√mn  + 5n
y–4 m–n

= y – 6√y  + 8 = 8m + 13√mn  + 5n
y–4 m–n

14. Selesaikan setiap yang berikut: TP 4 (a) √7x – 3 = 4

Solve each of the following: (√7x – 3)2 = 42

CONTOH 7x – 3 = 16

√2x + 1 = 3 7x = 19

Penyelesaian: x = 19
7
(√2x + 1 )2 = 32
2x + 1 = 9
2x = 8
x = 4

70

  Matematik Tambahan  Tingkatan 4  Bab 4 Indeks, Surd dan Logaritma 

(b) √2y + 1 = √5y – 11 (c) 3√2x – 1 = 3

(√2y + 1)2 = (√5y – 11)2 3√2x – 1 = 3
2y + 1 = 5y – 11
5y – 2y = 12 √2x – 1 = 3
3y = 12 3
y = 4
(√2x – 1)2 = 12

2x – 1 = 1

2x = 2

x = 1

(d) 5√m + 1 = 6 (e) √p  + 6 = p

(5√m + 1)2 = 62 √p  + 6 = p BAB 4

25(m + 1) = 36 (√p )2 = (p – 6)2

25m + 25 = 36 p = p2 – 12p + 36

25m = 11 p2 – 13p + 36 = 0

m = 11 (p – 4)(p – 9) = 0
25
(p – 4) = 0   atau p – 9 = 0

p = 4 p = 9

4.3 NOTHAukIMumBALSoAgNaritma

Laws of Logarithms

NOTA IMBASAN

Logaritma / Logarithms

71

  Matematik Tambahan  Tingkatan 4  Bab 4 Indeks, Surd dan Logaritma

15. Tukar setiap yang berikut dalam bentuk logaritma. TP 2

Change each of the following in logarithmic form.

CONTOH (a) 35 = 243 (b) 53 = 125

81 = 34  N = ax 243 = 35 125 = 53
log3 81 = 4  loga N = x log3 243 = 5 log5 125 = 3

(c) 7–2 = 1 (d) 102 = 100 (e) 9—12 = 3
49
100 = 102
log7 1 = 7–2 log10 100 = 2 1
49
3 = 92

BAB 4 log7 1 = –2 log9 3 = 1
49 2

16. Tukar setiap yang berikut dalam bentuk indeks. TP 2 (a) log2 32 = 5
25 = 32
Change each of the following in index form.
(d) log10 0.001 = –3
CONTOH 10–3 = 0.001

log2 8 = 3 Tip
23 = 8
Jika / If loga N = x,

maka / then ax = N

(b) log4 64 = 3 (c) log7 √7 = 1
43 = 64 2

7—12 = √7

17. Cari nilai bagi setiap yang berikut. TP 3

Find the value of each of the following.

CONTOH

(i) log5 √5 (ii) log3 1
Penyelesaian: 27

(i) Katakan log5 √5 = x (ii) Katakan log3 1 = y
27
Let Tukar kepada bentuk indeks. Let 1
   5x = √5  Change to index form. 27
5x = 5—21    Bandingkan indeksnya. 3y =
Compare index.
x = 1 = 1
2 33

= 3–3

y = –3

72

  Matematik Tambahan  Tingkatan 4  Bab 4 Indeks, Surd dan Logaritma 

(a) log4 256 (b) log6 1 (c) log3 81
36
Katakan log4 256 = x 1 Katakan log3 81 = x
Let 4x = 256 Katakan log6 36 = x Let (3)x = 81

4x = 44 Let 1
6x = 36
x = 4 3—21 x = 34

6x = 6–2 1  x = 4
2
x = –2 x = 8

(d) log10 0.01 (e) logm 1 (f) log—12 0.125

Katakan log10 0.01 = x Katakan logm 1 = x 0.125 =
Let 10x = 0.01 1 2 log —21


Let mx = 1 Katakan 1 x = x
Let 2 0.125
10x = 10–2 mx = m0

x = –2 x = 0 1 1 2x = 1 BAB 4
2 8

1 1 2x = 1 1 23
2 2

x = 3

18. Cari nilai x dalam setiap persamaan yang berikut. TP 3

Find the value of x in each of the following equations.

CONTOH

(i) log3 x = –4 (ii) logx 6 = 1
3
x = 3–4
x—13 = 6
x = 1 (x—13 )3 = 63
34

x = 1 x = 216
81

(a) log2 x = 6 (b) log3 x = 5 (c) log4 x = 1
x = 26 x = 35 2
x = 243
x = 64 4—12 = x
x = (22­)—21
x = 2

(d) logx 5 = 1 (e) log2x 1 = 1 (f) log—14 x = 2
3 8 2
x—13 = 5 1 41 22 = x
(2x)—21 = 1
1 2x —31 8
3 = 53 1 2 1 2 2x—21 2 = 1
12 x = 16
x = 125 8

2x = 1
64

x = 1
128

73

  Matematik Tambahan  Tingkatan 4  Bab 4 Indeks, Surd dan Logaritma

19. Ungkapkan setiap yang berikut sebagai satu logaritma tunggal. TP 4

Express each of the following as a single logarithm.

CONTOH (a) loga m + 3 loga n – loga m2n

3 loga x + 4 loga y – loga x2y = loga m + loga n3 – loga m2n

m × n3
m2n
1 2
= loga x3 + loga y 4 – loga x2y n loga x = loga x n = loga

x3 × y4 loga x + loga y = loga xy n2
x2y loga x – loga y = loga x m
1 2= loga   y = loga

(b) loga p – 3 loga q + 2 loga r (c) 1 loga m – 3 loga n + 2 loga q
2
= loga p – loga q3 + loga r2 = loga m—12 – loga n3 + loga q2

1 2 pr 2
= loga q3 1 2 q2m
= loga n3
BAB 4
20. Cari nilai bagi setiap yang berikut. TP 4
(a) log2 48 + log2 3 – log2 9
Find the value of each of the following.

CONTOH

2 log3 5 + log3 12 – 2 log3 10 = log2 1 48 × 3 2
9
= log3 52 + log3 12 – log3 102 n loga x = loga x n
= log2 16
1 25 × 12 2
= log3 100 loga x + loga y = loga xy = log2 24
loga x – loga y = loga x
= log3 3 y =4

=1

(b) log4 24 – log4 3 + log4 2 (c) log3 18 + log3 4 – 3 log3 2
4
= log3 18 + log3 4 – log3 23
18 × 4
24 × 2 23
3
1 2 = log4 = log3 1 2

4 = log3 9

= log4 64 = log3 32
= log4 43
=3 =2

1 2(d) log7 7 + log7 80 – log7 6 + log7 21 (e) 2 log2 6 – 1 log2 16 – 2 log2 3
40 2

7 × 80 × 21 62
40 (16)—12 × 32
1 2 = log7 1 2 = log2
6

= log7 49 = log2 1 36 2
= log7 72 4×9
=2
= log2 1
=0

74

  Matematik Tambahan  Tingkatan 4  Bab 4 Indeks, Surd dan Logaritma 

21. Selesaikan setiap yang berikut. TP 5

Solve each of the following.

CONTOH

Diberi log3 2 = 0.631 dan log3 5 = 1.465, cari nilai bagi setiap yang berikut.

Given log3 2 = 0.631 and log3 5 = 1.465, find the value of each of the following.

(i) log3 2 1 = log3 5 (ii) log3 200 = log3 (52 × 23) loga xy = loga x + loga y
2 2
x = log3 52 + log3 23
= log3 5 – log3 2 y = loga x – loga y
loga = 2 log3 5 + 3 log3 2 loga x n = n loga x

= 1.465 – 0.631 = 2(1.465) + 3(0.631)

= 0.834 = 4.823

Diberi log4 3 = 0.792 dan log4 5 = 1.161, cari nilai bagi setiap yang berikut. BAB 4

Given log4 3 = 0.792 and log4 5 = 1.161, find the value of each of the following.

(a) log4 1 2 = log4 5 (b) log4 20 = log4 20—12
3 3
1
= log4 5 – log4 3 = 2 log4 20
= 1.161 – 0.792
= 1 log4 (4 × 5)
= 0.369 2

= 1 (log4 4 + log4 5)
2

= 1 (1 + 1.161)
2

= 1.0805

(c) log4 3.75 = log4 3 3 (d) log4 45 = log4 (9 × 5)
4
15 = log4 (32 × 5)
4
= log4 = log4 32 + log4 5

= log4 1 3 × 5 2 = 2 log4 3 + log4 5
4
= 2(0.792) + 1.161

= log4 3 + log4 5 – log4 4 = 2.745
= 0.792 + 1.161 – 1

= 0.953

75

  Matematik Tambahan  Tingkatan 4  Bab 4 Indeks, Surd dan Logaritma

22. Cari nilai bagi setiap yang berikut. TP 3 (a) log4 7
= log10 7
Find the value of each of the following.
log10 4
CONTOH = 1.404

log5 21 Tukarkan asas 5 kepada asas 10. (d) log4 0.48
= log10 21 Change base 5 to base 10. = log10 0.48

log10 5 log10 8
= 1.892 = –0.3530

(b) log14 69 (c) log5 3
= log10 69 4

BAB 4 log10 14 = log10 3
= 1.604 4

log10 5

= –0.1787

23. Diberi log3 p = x, ungkapkan setiap yang berikut dalam sebutan x. TP 4

Given log3 p = x, express each of the following in terms of x.

CONTOH (a) log9 p = log3 p
log3 9
logp 3 = log3 3
log3 p = x
log3 32
= 1
log3 p loga a = 1 = x
2 log3 3

= 1 = x
x 2

(b) log9 27p = log3 27p (c) log9p 81 = log3 81
log3 9 log3 9p

= log3(33 × p) = log3 34
log3 32 log3 (32 × p)

= 3 log3 3 + log3 p = 4 log3 3
2 log3 3 2 log3 3 + log3 p

= 3+x = 4
2 2+x

76

  Matematik Tambahan  Tingkatan 4  Bab 4 Indeks, Surd dan Logaritma 

24. Selesaikan setiap persamaan yang berikut. TP 4

Solve each of the following equations.

CONTOH (a) 8x – 1 = 46

5x + 2 = 95 log10 8x – 1 = log10 46

log10 5x + 2 = log10 95 Ambil log10 pada kedua- (x – 1) log10 8 = log10 46
(x + 2) log10 5 = log10 95 dua belah persamaan.
Log
10 on both sides. x – 1 = log10 46
log10 8
x + 2 = log10 95
log10 5 x = 1.8412 + 1

x = 2.8295 – 2 x = 2.8412

x = 0.8295

(b) 5x – 2 = 4x + 1 (c) 2x3x = 5x + 1 BAB 4

log10 5x – 2 = log10 4x + 1 6x = 5x + 1

(x – 2) log10 5 = (x + 1) log10 4 x log10 6 = (x + 1) log10 5

x log10 5 – 2 log10 5 = x log10 4 + log10 4 x log10 6 = x log10 5 + log10 5

x log10 5 – x log10 4 = log10 4 + 2 log10 5 x log10 6 – x log10 5 = log10 5

x(log10 5 – log10 4) = log10 4 + 2 log10 5 x (log10 6 – x log10 5) = log10 5

x = log10 4 + 2 log10 5 x = log10 5
log10 5 – log10 4 log10 6 – log10 5

x = 20.64 x = 8.827

25. Selesaikan setiap persamaan yang berikut. TP 4

Solve each of the following equations.

CONTOH (a) logx 750 – 3 = logx 6

log3 4x = 3 log3 8 logx 750 – logx 6 = 3
4x = 83
logx 1 750 2 = 3
4x = 512 6

x = 512 logx 125 = 3
4 x3 = 125

x = 128 x3 = 53

x = 5

(b) 2 log7 (x – 2) = log7 25 (c) log2 4x = 6 – log2(5x – 2)

log7 (x – 2)2 = log7 25 log2 4x + log2(5x – 2) = 6
(x – 2)2 = 25 log2 4x(5x – 2) = 6

(x – 2)2 = 52 4x(5x – 2) = 26

x – 2 = 5 20x2 – 8x – 64 = 0

x = 7 5x2 – 2x – 16 = 0

(x – 2)(5x + 8) = 0

x= 2  atau  x = – 8   Jawapan tidak diterima kerana
5 logaritma bagi nombor negatif
tidak tertakrif.

77

  Matematik Tambahan  Tingkatan 4  Bab 4 Indeks, Surd dan Logaritma

(d) log5 (8x – 4) = 2 log5 3 + log5 4 (e) 2 log4 3 + log4(1 – x) – log4 3x = 1
2
log5 (8x – 4) = log5 32 + log5 4
log5 (8x – 4) = log5 (32 × 4) log4 32 + log4(1 – x) – log4 3x = 1
log5 (8x – 4) = log5 36 2
8x – 4 = 36
x = 5 log4 1 9 × (1 – x) 2 = 1
3x 2

9 × (1 – x) = 4—21
3x

9 – 9x = 2
3x
9 – 9x = 6x

15x = 9

x = 3
5

BAB 4

4.4 Aplikasi Indeks, Surd dan Logaritma

Applications of Indices, Surds and Logarithms

26. Selesaikan setiap masalah yang berikut. TP 5

Solve each of the following problem.

CONTOH

Diberi T = 2p M  , di mana T mewakili tempoh ayunan bandul dalam saat dan M ialah jisim bagi bandul
10

dalam kg. Cari jisim bandul apabila tempoh ayunan bandul ialah 1.14 saat.

Given T = 2p M , where T represents the period of the pendulum expressed in seconds and M represents the mass of the
10

pendulum in kg. Find the mass of the pendulum when the period of the pendulum is 1.14 seconds.

Penyelesaian:

T = 2p M 1 M 22 = 1 T 22 Apabila T = 1.14 s,
10 10 2p When
Maka, M = 10(1.14)2
M = T 1M0 = T2 Thus, 41 22 22
10 2p 4p2 7

M = 10T 2 = 0.33 kg
4p2

78

  Matematik Tambahan  Tingkatan 4  Bab 4 Indeks, Surd dan Logaritma 

(a) Diberi bahawa dua pemboleh ubah x dan y dihubungkan oleh persamaan

Given that two variables x and y are related by the equation

y = 9 (x – 0.4)2
4
Cari nilai x apabila y = 1.9.

Find the value of x when y = 1.9.

y = 9  (x – 0.4)2­ x= 2√y + 0.4
4 3

(x – 0.4)2 = 4y Apabila / When y = 1.9,
9
2√1.9
x – 0.4 = 4y Maka / Thus,  x = 3 + 0.4
9
= 1.319

BAB 4

(b) Rajah menunjukkan sebuah segi empat tepat PQRS dengan panjang sisi P (1 + ͱ⒓5) cm Q
S R
ialah (1 + √5) cm dan luas segi empat ialah √80 cm2. Hitungkan lebar,
dalam cm, segi empat tepat itu. Ungkapkan jawapan anda dalam bentuk

m + n√5 di mana m dan n ialah integer.
The diagram shows a rectangle PQRS with the length of (1 + √5) cm and the area of
the rectangle is √80 cm2. Calculate the width, in cm, of the rectangle. Express your
answer in the form of m + n√5 where m and n are integers.

Luas = Panjang × Lebar = √16 × 5 1 – √5 = 4√5 – 4√5 √5
  Area = Length × Width (1 + √5) (1 – √5) 1–5

√80 = (1 + √5) × Lebar = 4√5 1 – √5 = 4√5 – 20
(1 + √5) (1 – √5) –4
Lebar  = √80
Width (1 + √5) = (5 – √5) cm

= √80 1 – √5
(1 + √5) (1 – √5)

79

  Matematik Tambahan  Tingkatan 4  Bab 4 Indeks, Surd dan Logaritma 4

PRAKTIS SPM

Kertas 1 3. Selesaikan persamaan:

1. Diberi 27k – 2 = 1, ungkapkan p dalam sebutan k. SPM Solve the equation:
243p + 1
SPM 2013 log4 3 + log4 (2x – 1) = 2

2017 Given 27k – 2 = 1, express p in terms of k. log4 3 + log4 (2x – 1) = 2
243p + 1 log4 3(2x – 1) = 2
3(2x – 1) = 42
22473kp–+21 = 1
6x – 3 = 16
27k – 2 = 243p + 1
6x = 19
(33)k – 2 = (35)p + 1
x = 19
33k – 6 = 35p + 5 6

3k – 6 = 5p + 5

5p = 3k – 11 4. Diberi log3 x = p dan log3 y = q, ungkapkan

p = 3k – 11
5
log3 81x2 dalam sebutan p dan q.
y
BAB 4
Given log3 x = p and log3 y = q, express log3 81x2 in
terms of p and q. y

81x2 81x2 1
1 2log3 y = 2
log3
2. Selesaikan persamaan: y

SPM Solve the equation: 34 x2 1
1 2 2
2017 logh 864 – log√h 2h = 1 = log3
y

logh 864 – log√h 2h = 1 1 2 32 x
= log3 y 21
logh 2h
logh 864 – logh √h = 1 = log3 32 + log3 x – log3 y 21

logh 864 – logh 2h = 1 = 2 log3 3+ log3 x – 1 log3 y
2
1

logh 864 – lloogghh h2 = 1 =2 + p– q
2h 2

1 1 2
2
logh 864 – 2 logh 2h = 1
logh 864 – logh (2h)2 = 1 5. Maklumat berikut adalah berkaitan dengan
864 SPM hukum indeks:
logh 4h2 = 1
2019 The following information is on the laws of indices.

864 = h (am)10 = 3√a  × 3√a  × 3√a  × … × 3√a 
4h2 p kali / p times

216 = h3 dengan keadaan m dan p ialah pemalar.

63 = h3 where m and p are constants.

h = 6 Nyatakan nilai m dan nilai p.

State the value of m and of p.

1

(am)10 = (a 3 )p

\  m = 1 ,  p = 10
3

80

  Matematik Tambahan  Tingkatan 4  Bab 4 Indeks, Surd dan Logaritma 

6. Diberi logn K = 3x, ungkapkan dalam sebutan x, 2. Selesaikan persamaan:
SPM Given logn K = 3x, express in terms of x,
2019 Solve the equation:

1 log4(7x + 3) – log4 x2 + log16 x2 = 3
K
(a) logn log4(7x + 3) – log4 x2 + log4 x2 = 3
log4 16
(b) log√n Kn4 log4 x2
log4 42
1 log4(7x + 3) – log4 x2 + = 3
K
(a) logn = logn K–1 2 log4(7x + 3) – 2 log4 x2 + log4 x2 = 6
= –logn K

= –3x (7x + 3)2x2
x4
(b) log√n Kn4 = log√n K + log√n n4 log4 = 6

= logn K + 4 logn n (7x + 3)2x2 = 46
x4
1 1 (7x + 3)2
x2
logn n2 logn n2 = 4 096

= 3x + 4 (7x + 3)2 = 4 096x2
1 1
(7x + 3)2 = (64x)2
22
7x + 3 = 64x
= 6x + 8
x = 1
19 BAB 4

Kertas 2 3. D(Gaiiv)b eenudrenilonlggogk9gaax9npx=ka=masnamasnx3ddy.alnodgal3noyg3=yxyn=,danl,am

1. Diberi bahawa h = 5x dan k = 5y. KBAT

SPM It is given that h = 5x and k = 5y. bentuk indeks
2014
(a) Ungkapkan 25x + y dalam sebutan h dan k.
125y express xy and x in index form with base 3.
Express 25x + y in terms of h and k. y

125y (b) Seterusnya, hitung nilai m dan nilai n
x
(b) Cari log25 125h dalam sebutan x dan y. diberikan xy = 243 dan y = 27.
k2
Hence, calculate the value of m and of n given
125h x
Find log25 k 2 in terms of x and y. xy = 243 and y = 27.

(a) 25x + y = 52(x + y) (a) log9 x = m xy = 32m × 3n
125y 53y x = 9m
= 32m = 32m + n
log3 y = n
= 52x + 2y – 3y y = 3n x = 32m
y 3n
= 52x – y

= (5x)2 = 32m – n
5y
(b) xy = 243
h2
= k 32m + n = 35

2m + n = 5 …… 1

(b) log25 125h = log25 125 + log25 h – log25 k2 x = 27
k2 y

= log5 53 + log5 h – log5 k2 32m – n = 33
log5 52 log5 52 log5 52
2m – n = 3 ……… 2

= log5 53 + log5 5x – log5 52y 1 + 2, 4m = 8
log5 52
m = 2

= 3 + x– 2y Daripada 1, 2(2) + n = 5 Praktis
2 SPM
n = 1 Ekstra

81

  Matematik Tambahan  Tingkatan 4  Bab 4 Indeks, Surd dan Logaritma KBAT KBAT

Sudut Ekstra

1. Farid telah berkhidmat di sebuah syarikat dengan gaji tahunannya diberi oleh persamaan G = 18 000(1.05)t – 1,
dengan keadaan G ialah gaji tahunan dan t ialah bilangan tahun dia berkhidmat.Cari nilai minimum bagi t
supaya gaji tahunannya akan melebihi RM60 000.

Farid started working for a company with an annual salary given by an equation G = 18 000(1.05)t – 1 such that G is the
annual salary and t are the number of years he is working in the company. Find the minimum value of t such that his
annual salary will exceed RM60 000.

18 000(1.05)t – 1 . 60 000

(1.05)t – 1 . 60 000
18 000

(1.05)t – 1 . 10
3
1 2 (t – 1) log10 1.05 . log10 10
3

1 2 log10 10
t – 1 . 3
BAB 4 log10 1.05

t – 1 . 24.68

t . 25.68

t = 26

2. Suatu syarikat pelaburan menawarkan suatu pulangan r% setahun. Sejumlah wang RMp dilaburkan.

Jumlah pelaburan selepas suatu tempoh n tahun diberikan oleh RMp11 + r 2n tertakluk kepada syarat
100
iaitu pelaburan itu tidak dikeluarkan. Swee Hoe melaburkan RM30 000 dan pulangan ialah 6% setahun.

An investment company offers a return of r% per annum. A sum of money RMp is invested. The total investment after a

1 2period of n years is given by RMp1+ r n subject to the condition that the investment is not withdrawn. Swee Hoe
100

invested RM30 000 and the return is 6% per annum.

(a) Berapakah jumlah pelaburan selepas 5 tahun?

How much is the investment after 5 years?

(b) Selepas berapa tahunkah jumlah pelaburan akan melebihi RM100 000 untuk kali pertama?

After how many years will the total investment exceed RM100 000 for the first time?

(a) Jumlah pelaburan = p11 + 2r rt (b) Jumlah pelaburan > 100 000

100 6 25 30 000 (1.06)n > 100 000
00011 100
= 30 + 1 2n 10
log10 1.06 > log10 3

= 30 000 (1.06)5 1 2n > log10 10
3
n >
= RM40 146.77 2lo0g.61061.06

n = 21 tahun

Kuiz 4

82

BAB Janjang

5 Progressions

4.1 Janjang Aritmetik
5.1
Arithmetic Progressions

NOTA IMBASAN

NOTA IMBASAN

Janjang aritmetik
Arithmetic progressions

1. Tentukan sama ada setiap yang berikut ialah janjang aritmetik atau bukan.   TP 1

Determine whether each of the following is an arithmetic progression.

CONTOH 1 CONTOH 2

19,15,11,7,… log10 5, log10 10, log10 20, log10 80

Penyelesaian: Penyelesaian: 10
T2 − T1 = 15 − 19 = – 4 5
T3 − T2 = 11 − 15 = –4 T2 − T1 = log10 10 − log10 5 = log10 = log10 2
T4 − T3 = 7 − 11 = –4
T3 − T2 = log10 20 − log10 10 = log10 20 = log10 2
d = –4 ialah pemalar/ is a constant. 10
\ Ini ialah janjang aritmetik.
T4 − T3 = log10 80 − log10 20 = log10 80 = log10 4
\  This is an arithmetic progression. 20

d bukan satu pemalar/ d is not a constant.

Bukan suatu janjang aritmetik.

Not an arithmetic progression.

83

  Matematik Tambahan  Tingkatan 4  Bab 5 Janjang

(a) 3 , 1 1 , 9 , 3, … (b) log5 4, log5 12, log5 36, log5 108, (c) 3p – 2, 2p – 1, p, 5 + p, …
4 2 4 …

T2 − T1 = 1 1 − 3 = 3 T2 − T1 = log5 12 = log5 3 T2 − T1 = 2p − 1 – (3p – 2) = 1 – p
2 4 4 4 T3 − T2 = p − (2p – 1) = 1 – p
T4 − T3 = 5 + p − p = 5
T3 − T2 = 9 − 1 1 = 3 T3 − T2 = log5 36 = log5 3
4 2 4 12 d bukan satu pemalar
Bukan suatu janjang aritmetik
T4 − T3 = 3 − 9 = 3 T4 − T3 = log5 108 = log5 3
4 4 36 d is not a constant

d= 3 (pemalar / constant) d = log5 3 (pemalar / constant) Not an arithmetic progression
4 Suatu janjang aritmetik

Suatu janjang aritmetik An arithmetic progression

An arithmetic progression

2. Nyatakan sebutan pertama dan beza sepunya bagi setiap janjang aritmetik yang berikut. TP 2

State the first term and the common difference for each of the following arithmetic progression.

BAB 5 CONTOH 1 CONTOH 2 CONTOH 3

−3, 5, 13, 21, … 17, 12, 7, 2, … x + 3, 2x + 5, 3x + 7, 4x + 9, …

Penyelesaian: Penyelesaian: Penyelesaian:
a = −3 a = 17 a = x + 3
d = 5 − (−3) = 8 d = 12 − 17 = –5 d = 2x + 5 – (x + 3) = x + 2

(a) –6, 3, 12, 21, … (b) 11, 8 1 ,  6, 3 1 , … (c) 3x + 7, 4x + 5, 5x + 3, 6x + 1, …
2 2
a = –6 a = 3x + 7
d = 3 − (–6) = 9 a = 11 d = 5x + 3 – (4x + 5) = x – 2

d= 6 − 8 1 = –2 1
2 2

3. Tentukan nilai sebutan tertentu bagi janjang aritmetik yang berikut. TP 3

Determine the value of the specific term for the arithmetic progression.

CONTOH 1 CONTOH 2

11, 7, 3, … log 3, log 6, log 12,…, …
Cari / Find  T9.
Penyelesaian: Cari / Find  T7.
a = T1 = 11 Penyelesaian:
d = T2 – T1 = 7 – 11 = – 4
Tn = a + (n – 1)d a = T1 = log 3 – log 6 = log 12 = log 2
T9 = 11 + (9 – 1)(– 4) d = T3 – T2 = log 12 6
= 11 – 32 Tn = a + (n – 1)d
= –21
T7 = log 3 + (7 – 1) log 2
= log 3(26)

= log (3 × 64)

= log 192

84

  Matematik Tambahan  Tingkatan 4  Bab 5 Janjang 

(a) −8, −2, 4, 10, … (b) log 10, log 100, log 1000, …
Cari / Find T12
Cari / Find T8
a = −8 
d = 4 − (−2) = 6 a = log 10 100
T12 = –8 + 11(6) 10
= 58 d = log 100 – log 10 = log = log 10
T8 = log10 10 + 7(log 10)

= log 10(107)

= 8 log 10

4. Hitungkan bilangan sebutan bagi setiap janjang aritmetik yang berikut. TP 3

Calculate the number of terms in each of the following arithmetic progressions.

CONTOH 1 CONTOH 2 BAB 5

4.1, 4.7, 5.3, 5.9, …, 11.3 3y, –y, –5y, …, –65y

Penyelesaian: Penyelesaian:
a = 4.1,  d = 4.7 − 4.1 = 0.6 a = 3y,  d = –y − 3y = –4y
Tn = a + (n − 1)d Tn = a + (n − 1)d
= 4.1 + (n − 1)(0.6) = 3y + (n − 1)(–4y)
= 0.6n + 3.5 = 7y − 4ny
0.6n + 3.5 = 11.3
0.6n = 7.8 –4ny + 7y = –65y
n = 13 –4ny = –72y
n = 18

(a) −11, −7, −3, …, 125 (b) 17, 13, 9, …, –31

a = −11 , d = −7 − (−11) = 4 a = 17, d = 13 − 17 = −4
Tn = a + (n − 1)d Tn = a + (n − 1)d
= –11 + (n – 1)(4) = 17 + (n − 1)( −4)
= 4n − 15 = −4n + 21

4n – 15 = 125 –4n + 21 = –31
4n = 140 4n = 52
n = 35 n = 13

(c) 1 , 2 , 3 , …, 14 (d) 3m, m, –m, –3m, …, −13m
5 5 5 5
1 2 1 1 a = 3m , d = m − 3m = −2m
a = 5 ,d = 5 − 5 = 5 Tn = a + (n − 1)d
= 3m + (n − 1)( −2m)
Tn = a + (n − 1)d 1 = 5m – 2nm
= 1 5
5 + (n – 1)( ) 5m – 2mn = –13m
2nm = 18m
= 1 n 2n = 18
5 n = 9
1 14
5 n = 5

  n = 14

85

  Matematik Tambahan  Tingkatan 4  Bab 5 Janjang

5. Tentukan hasil tambah n sebutan pertama bagi suatu janjang aritmetik. TP 3

Determine the sum of the first n terms in an arithmetic progression.

CONTOH 1 CONTOH 2

3, 5, 7, 9, … (12 sebutan yang pertama) 5, 2, –1, –4, … (10 sebutan yang pertama)
(first 10 terms)
(first 12 terms)

Penyelesaian: Penyelesaian:

a = 3, d = 5 − 3 = 2 a = 5,  d = 2 − 5 = −3

S12 = n [2a + (n – 1)d] Sn = n [2a + (n – 1)d]
2 2

= 122[2(3) + 11(2)] S10 = 10 [2(5) + 9(–3)]
2
= 6(6 + 22)
= 5(10 – 27)
= 168 = –85

(a) −15, −11, −7, −3, … (16 sebutan yang pertama) (b) 19, 14, 9, 4, … (14 sebutan yang pertama)
(first 16 terms)
(first 14 terms)

BAB 5 a = −15,  d = −11 − (−15) = 4 a = 19,  d = 14 − 19 = –5

S16 = n [2a + (n – 1)d] S14 = n [2a + (n – 1)d]
2 2

= 16 [2(–15) + 15(4)] = 124[2(19) + 13(–5)]
2

= 8(–30 + 60) = 7(38 – 65)

= 240 = –189

6. Tentukan hasil tambah sebutan tertentu yang berturutan bagi suatu janjang aritmetik. TP 3

Determine the sum of specific number of consecutive terms in an arithmetic progression.

CONTOH 1 CONTOH 2

23, 17, 11, …., –67 –6, –4, –2, 0, …., 16

Penyelesaian: Penyelesaian:
a = 23,  d = 17 − 23 = −6, l = –67 a = –6,  d = –4 − (–6) = 2, l = 16
Tn = a + (n – 1)d Tn = a + (n – 1)d
= 23 + (n – 1)(–6) = –6 + (n – 1)(2)
= 29 – 6n = 2n – 8

29 – 6n = –67 2n – 8 = 16

6n = 96 2n = 24

n = 16 n = 12
n n
S16 = 2 [a + l] S12 = 2 [a + l]

= 126[23 + (–67)] = 122[–6 + 16]

= 8(–44) = 6(10)

= –352 = 60

86

  Matematik Tambahan  Tingkatan 4  Bab 5 Janjang 

(a) −15, −12, −9, …, 30 (b) 9, 4, –1, …, –86

a = −15,  d = −12 − (−15) = 3, l = 30 a = 9,  d = 4 − 9 = –5, l = –86
Tn = a + (n – 1)d Tn = a + (n – 1)d
= –15 + (n – 1)(3) = 9 + (n – 1)(–5)
= 3n – 18 = 14 – 5n

3n – 18 = 30 14 – 5n = –86

3n = 48 5n = 100

n = 16 n = 20
n n
S16 = 2 [a + l] S20 = 2 [a + l]

= 16 [–15 + 30] = 20 [9 – 86]
2 2

= 120 = –770

7. Hitung hasil tambah sebutan tertentu yang berturutan bagi suatu janjang aritmetik. TP 4

Calculate the sum of specific number of consecutive terms in an arithmetic progression.

CONTOH 1 BAB 5

3, 10, 17, …. (sebutan ke-7 hingga sebutan ke-12) Tip

(7th term to 12th term) S12 = T1 + T2 + T3 + T4 + T5 + T6 + T7 + T8 + T9 + T10 + T11 + T12
S6 = T1 + T2 + T3 + T4 + T5 + T6
Penyelesaian: \ S12 – S6 = T7 + T8 + T9 + T10 + T11 + T12
{Hasil tambah T7 hingga T12}
{Sum of T7 to T12}

Kaedah 1/ Method 1 Kaedah 2/ Method 2
a = 3,  d = 10 − 3 = 7 T7 = 3 + 6(7) = 45
T12 = 3 + 11(7) = 80
Hasil tambah T7 hingga T12/ Sum of T7 to T12 45, …, 80 (6 sebutan/ terms)
= S12 – S6
Hasil tambah T7 hingga T12/ Sum of T7 to T12
n
= 122[2(3) + 11(7)] – 6 [2(3) + 5(7)] S6 = 2 [a + l]
2
6
= 6(83) – 3(41) = 2 [45 + 80]

= 375 = 375



(a) −22, −16, −10, … (sebutan ke-5 hingga sebutan ke-14) (b) –1, 4, 9, … (sebutan ke-4 hingga sebutan ke-12)
(5th term to 14th term)
(4th term to 12th term)

a = −22,  d = −16 − (−22) = 6 a = –1,  d = 4 − (–1) = 5

Hasil tambah T5 hingga T14 Hasil tambah T4 hingga T12

Sum of T5 to T14 Sum of T4 to T12

= S14 – S4 = S12 – S3

= 14 [2(–22) + 13(6)] – 4 [2(–22) + 3(6)] = 12 [2(–1) + 11(5)] – 3 [2(–1) + 2(5)]
2 2 2 2

= 7(34) – 2(–26) = 6(53) – 12

= 290 = 306

87

  Matematik Tambahan  Tingkatan 4  Bab 5 Janjang

(c) 6.7, 5.3, 3.9, … (sebutan ke-6 hingga sebutan ke-10) (d) 5, 92 , 4, … (sebutan ke-9 hingga sebutan ke-16)

(6th term to 10th term) (9th term to 16th term)

a = 6.7,  d = 5.3 − 6.7 = −1.4 a = 5,  d = 9 − 5 = –0.5
2

Hasil tambah T6 hingga T10 Hasil tambah T9 hingga T16

Sum of T6 to T10 Sum of T9 to T16

= S10 – S5 = S16 – S8

= 10 [2(6.7) + 9(–1.4)] – 5 [2(6.7) + 4(–1.4)] = 126 [2(5) + 15(–0.5)] – 8 [2(5) + 7(–0.5)]
2 2 2

= 5(0.8) – 19.5 = 8(2.5) – 4(6.5)

= –15.5 = –6

8. Selesaikan masalah berikut yang melibatkan janjang aritmetik. TP 5

Solve the following problems involving arithmetic progressions.

CONTOH 1 (a) Diberi p, 3p – 1 dan 4p adalah tiga sebutan

BAB 5 Diberi x + 3, 2x – 3 dan x – 5 adalah tiga sebutan berturut-turut bagi janjang aritmetik. Cari nilai

berturut-turut bagi janjang aritmetik. Cari nilai bagi p.

bagi  x. Given that p, 3p – 1 and 4p are three consecutive terms
of an arithmetic progression. Find the value of p.
Given that x + 3, 2x – 3 and x – 5 are three consecutive
terms of an arithmetic progressions. Find the value of x. 4p – (3p – 1) = 3p – 1 – p

Penyelesaian: p + 1 = 2p – 1
2 x – 3 – (x + 3) = x – 5 – (2x – 3)
2x – 3 – x – 3 = x – 5 – 2x + 3 p = 2
x – 6 = –x – 2
2x = 4 Tip
x = 2
d = Tn – Tn – 1

CONTOH 2 (b) Sebutan ke-n, Tn suatu janjang aritmetik diberi
sebagai Tn = 5n – 2. Cari hasil tambah 7 sebutan
Sebutan ke-n, Tn suatu janjang aritmetik diberi pertama janjang itu.
sebagai Tn = 4n + 7. Cari hasil tambah 5 sebutan
pertama janjang itu. The nth ter m– ,2T. nFionfdatnhearsiuthmmoetficthperfoirgsrte7ssitoenrmiss given
as Tn = 5n of the

The nth term, TFn inofd an arithmetic progression is given as progression.
pTnro=gre4snsio+n. 7. the sum of the first 5 terms of the

Penyelesaian: Tn = 5n – 2
a = T1 = 5(1) – 2 = 3
Tn = 4n + 7
a = T1 = 4(1) + 7 = 11 T2 = 5(2) – 2 = 8
T2 = 4(2) + 7 = 15 d = T2 – T1 = 8 – 3 = 5
d = T2 – T1 = 15 – 11 = 4
S7 = 7 [2(3) + 6(5)]
2

S5 = 5 [2(11) + 4(4)] = 126
2

= 95

88

  Matematik Tambahan  Tingkatan 4  Bab 5 Janjang 

CONTOH 3 (c) Diberi hasil tambah n sebutan pertama bagi

Diberi hasil tambah n sebutan pertama bagi janjang janjang aritmetik –22, –16, –10, … ialah 132.
aritmetik 25, 18, 11, … ialah 4. Cari nilai bagi  n.
Cari nilai bagi n.
Given the sum of the first n term of the arithmetic progression
25, 18, 11, … is 4. Find the value of n. Given the sum of the first n term of the arithmetic
progression –22, –16, –10, … is 132. Find the value
of n.

Penyelesaian: a = –22,  d = –16 – (–22) = 6

a = 25,  d = 18 – 25 = –7 Sn = n [2(–22) + (n – 1)(6)]
n 2
Sn = 2 [2(25) + (n – 1)(–7)] 2n [–44 – 6 + 6n] = 132

n2 [50 + 7 – 7n] = 4 6n2 – 50n – 264 = 0

7n2 – 57n + 8 = 0 3n2 – 25n – 132 = 0

(7n – 1)(n – 8) = 0 (3n + 11)(n – 12) = 0

n= 1 (tidak diterima/ not accepted),   n=8 n = – 11   (tidak diterima/ not accepted), n = 12
7 3

CONTOH 4 (d) Dalam suatu janjang arimetrik, sebutan ke-5

Dalam suatu janjang arimetrik, sebutan ke-6 ialah ialah 13 dan sebutan ke-12 ialah 27. Cari sebutan BAB 5
28 dan sebutan ke-10 ialah 16. Cari sebutan pertama
dan beza sepunya. pertama dan beza sepunya.

In an arithmetic progression, the 6th term is 28 and the 10th In an arithmetic progression, the 5th term is 13 and the
term is 16. Find the first term and the common difference. 12th term is 27. Find the first term and the common
difference.

Penyelesaian:   T5 = 13 = a + 4d …… (i)
T6 = 28 = a + 5d …… (i) T12 = 27 = a + 11d ……(ii)
T10 = 16 = a + 9d …… (ii) (ii) –(i): 7d = 14
(ii) – (i): 4d = –12
d = –3 d = 2
28 = a + 5(–3)
a = 43 Tip 13 = a + 4(2)

Tn = Sn – Sn – 1 a = 5

CONTOH 5 (e) Hasil tambah n sebutan pertama suatu janjang
5
Hasil tambah n sebutan pertama suatu janjang aritmetik diberi oleh Sn = 2 n2 – 9n. Hitung
aritmetik diberi oleh Sn = 3n2 + n. Hitung sebutan sebutan ke-5.
ke-6.
The sum of the first n term of an arithmetic progression
The sum of the first n term of an arithmetic progression is 5
given by Sn = 3n2 + n. Calculate the 6th term. is given by Sn = 2 n2 – 9n. Calculate the 5th term.

Penyelesaian: T5 = S5 – S4
T6 = S6 – S5
= [3(6)2 + 6] – [3(5)2 + 5] 3 4 3 4 = 5 – – 5 –
= 114 – 80 2 (5)2 9(5) 2 (4)2 9(4)
= 34
3 4 = 125 – 45 – [40 – 36]
2

= 35 – 4
2

= 13.5

89