Module & More Add Math Tg 5

BAB 2   Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan

23. Selesaikan masalah yang melibatkan kadar perubahan bagi kuantiti yang terhubung dan mentafsir penyelesaian
tersebut.

Solve the problems involving the rate of change of related quantities and interpret the answers. TP 5

CONTOH

Setelah dipanaskan, panjang sisi x cm bagi dV = 3x2 dx 
sebuah kubus bertambah dengan kadar malar dt dt
0.002 cm s−1. Cari
= 3(5)2(0.002)
Once heated, the length of the side of x cm of a cube
increases with a constant rate of 0.002 cm s−1. Find = 0.15 cm3 s−1

(i) kadar perubahan isi padunya jika x = 5 cm, Isi padu bertambah dengan kadar 0.15 cm3 s−1.

the rate of change in the volume if x = 5 cm, The volume increases with a rate of 0.15 cm3 s−1.

(ii) kadar perubahan dalam luas permukaan (ii) Katakan luas permukaan ialah L.
kubus jika x = 6 cm.
Let the surface area is L.
the rate of change in the surface area of cube if
x = 6 cm. L = 6x2

dL = 12x
dx
Penyelesaian:
dL = dL × dx
(i) Katakan V ialah isi padu kubus. dt dx dt
Let V is the volume of cube. dx
Diberi/Given x = 6 cm dan/and dt = 0.002
Maka/Hence V = x3

dV = 3x2 x dL = 12(6)(0.002)
dx dt
= 0.144 cm2 s−1
dV dV dx x
dt = dx × dt x Luas permukaan bertambah dengan kadar

Diberi/Given dx = 0.002 cm s−1, x = 5 cm 0.144 cm2 s−1.
dt
The surface area increases with a rate of
0.144 cm2 s−1.

(a) Sebuah kon tegak yang terbalik mempunyai sudut separa 30°. Jika kon itu diisi dengan minyak
hingga penuh dan minyak kemudiannya dibiar mengalir keluar dari muncung kon dengan kadar
4 cm3 s−1. Cari kadar perubahan aras minyak di dalam kon yang menurun ketika tinggi aras minyak
itu ialah 4 cm.

An inverted cone has a semi angle of 30°. If the cone is filled fully with oil and then the oil is allowed to drip out
from the tip of the cone at a rate of 4 cm3 s−1. Find the rate of change of the height of oil in the cone that is decreasing
when the height of the oil is 4 cm.

tan 30° = r = 1 Jika h = 4 cm, dV = –4 cm3 s–1
h 3 dt

r = h –4 = 1 π(4)2 dh
3 dt
r
1 3 – 43π = dh h 30Њ
3 dt
 Isipadu V = 1 πr2h
3 πh 2h ∴ dh = –3 cm s–1
Volume dt 4π
3
= h menyusut

V = 1 πh3 h decreases
9

dV = 1 πh2
dh 3

dV = 1 πh2 dh
dt 3 dt

48

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan 

(b) Luas sebuah bulatan bertambah dengan kadar (c) Isi padu, V cm3 sebuah silinder menyusut BAB 2
dengan kadar 1.2 cm3 s−1 dan tingginya h cm
π cm2 s−1. Tentukan kadar perubahan jejarinya sentiasa dua kali panjang jejari, r cm tapak
silinder. Cari kadar perubahan jejari apabila r
apabila jejarinya ialah 3.5 cm. ialah 7 cm.

The area of a circle increases at a rate of π cm2 s−1. The volume, V cm3, of a cylinder decreases at a rate of
Determine the rate of change of its radius when the 1.2 cm3 s−1 and its height of h cm is always twice of
radius is 3.5 cm. the radius, r cm of the base of the cylinder. Find the
rate of change of the radius when the r is 7 cm.

L = πr2 h = 2r

dL = 2πr V = πr2h r
dr h
= πr2(2r)

dL = 2πr dr V = 2πr3
dt dt
dV = 6πr2
2πr dr  dr
π = dt
dV dr
dr dt = 6πr2 dt
dt
1 = 2(3.5) dr
dt
dr 1 –1.2 = 6π(7)2
dt 7
= cm s–1 dr –1.2
dt 6π(7)2
Jejari bertambah = = –0.0013 cm s–1

The redius increases Jejari menyusut

The radius decreases

24. Tentukan perubahan kecil dan penghampiran suatu kuantiti bagi setiap yang berikut.

Determine the small change and the approximated quantity for each of the following. TP 4

CONTOH

Diberi y = x2 + x, cari tokokan kecil dalam y apabila x berubah daripada 2 kepada 1.98. Seterusnya, cari
nilai baharu y.

Given that y = x2 + x, find the small change in y when x changes from 2 to 1.98. Then, find the new value of y.

Penyelesaian:

Diberi/Given y = x2 + x ≈ (2x + 1)δx
≈ (2(2) + 1)(−0.02)
dy = 2x + 1 ≈ −0.1
dx Jadi, y menyusut sebanyak 0.1.

x berubah daripada 2 kepada 1.98. So, y decreases 0.1.

x changes from 2 to 1.98. Apabila/When x = 2, y = 22 + 2 = 6
Maka, nilai baharu y
Jadi/So δx = 1.98 – 2 = −0.02
Hence, the new value of y
δy ≈ dy
δx dx = y + δy
= 6 + (−0.1) = 5.9
δy ≈ dy ·δx
dx

49

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan

(a) Diberi y = 4x − 3, cari tokokan kecil dalam y apabila x berubah daripada 5 kepada 5.01. Seterusnya,
cari nilai baharu y.

Given that y = 4x − 3, find the small change in y when x changes from 5 to 5.01. Then, find the new value of y.

BAB 2 Diberi y = 4x − 3 ≈ 4 δx
≈ 4 (0.01)
dy =4 ≈ 0.04
dx Jadi, y bertambah sebanyak 0.04.

x berubah daripada 5 kepada 5.01 So, y increases 0.04.
x changes from 5 to 5.01
Apabila x = 5, y = 17
Jadi/So δx = 5.01 – 5 = 0.01 Maka, nilai baharu y

δδxy ≈ dy Hence, the new value of y
dx
= y + δy
δy ≈ dy · δx = 17 + (0.04)
dx = 17.04

(b) Diberi y = 9 , cari dy . Seterusnya, cari nilai (c) Diberi y = x2(x + 2), cari tokokan kecil dalam y
x2 dx apabila x berubah daripada 3 kepada 2.99.

hampir bagi 4 . Beri jawapan betul Given that y = x2(x + 2), find the small change in y
(2.98)2 when x changes from 3 to 2.99.

kepada dua angka bererti. Diberi y = x2(x + 2)

Given that y = 9 , find dy . Then, find the dy
x2 dx dx
4 . Give your answer = 3x2 + 4x
approximate value for
(2.98)2 x berubah daripada 3 ke 2.99.

correct to two significant figures. x changes from 3 to 2.99.

Diberi y = 9 Jadi, δx = 2.99 – 3 = −0.01
x2
= – 1x 83 δy ≈ dy
dy δx dx
dx
δy ≈ dy · δx
x berubah daripada 3 kepada 2.98. dx

x changes from 3 to 2.98. ≈ (3x2 + 4x)δx
≈ [3(3)2 + 4(3)](−0.01)
Jadi, δx = 2.98 − 3 = −0.02 ≈ −0.39
Jadi, y menyusut sebanyak 0.39.
δy ≈ dy
δx dx so, y decreases 0.39.

δy ≈ dy · δx
dx
–– 1(x13 83 8)3·(δ−x0.02)
≈
≈

≈ 0.013

Jadi, y bertambah sebanyak 0.013.

So, y increases 0.013.

Apabila x = 3, y = 1

Maka, nilai baharu y = y + δy

Hence, the new value of y = y + δy

= 1 + 0.013

= 1.013

50

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan 

25. Selesaikan setiap yang berikut.

Solve each of the following. TP 4

CONTOH (a) Sebuah silinder tanpa penutup mempunyai BAB 2

Jejari sebuah sfera bertambah daripada 3 cm tinggi 8 cm dan jejari j cm. Jika luas permukaan

kepada 3.01 cm. Cari perubahan kecil bagi isi silinder itu ialah L cm2,

padunya secara penghampiran A cylinder without a cover has a height of 8 cm and
a radius of j cm. If the surface area of the cylinder is
The radius of a sphere increases from 3 cm to 3.01 cm. L cm2,
Find the approximate change in its volume.
(i) tunjukkan bahawa L = π(j2 + 16j).
Penyelesaian: 4
3 show that L = π(j2 + 16j).
Isi padu sfera, V= πr3, dengan r ialah jejari sfera.
(ii) Seterusnya, cari perubahan kecil bagi
Volume of sphere, V = 4 πr3, where r is the radius of sphere.
3 luasnya jika jejari silinder itu menyusut
dV
dr = 4πr 2 daripada 4 cm kepada 3.98 cm.

Then, find the approximate change in the area if
the radius of the cylinder decreases from 4 cm
to 3.98 cm.

Diberi jejari sfera bertambah daripada 3 ke 3.01 cm. (i) L = 2πj × 8 + πj2
Given that the radius of sphere increases from 3 to
3.01 cm. = π[16j + j2] 8 cm
j
Maka/Hence (ii) δj = 3.98 – 4 = –0.02 cm

δr = 3.01 – 3 = 0.01 cm dL = 16π + 2πj
dj
δV ≈ dV
δr dr Apabila j = 4 cm dan δj = –0.02

δV ≈ dV · δr δL = dL . δj
dr dj

≈ (4πr2)δr = [16π + 2π(4)])(–0.02)

≈ 4π(3)2(0.01) = 0.36π cm3 = 24π(–0.02)

Jadi, isi padu sfera bertambah sebanyak 0.36π cm3. = –0.48π cm2

So, the volume of sphere increases 0.36π cm3. Luas menyusut sebanyak 0.48π cm2
The area decreases 0.48π cm2

(b) Isi padu, V cm3, bagi suatu bekas diberi oleh (c) Tempoh ayunan, T s, satu bandul yang
mempunyai panjang L cm diberi oleh
V = x(6x − 4), dengan keadaan x cm ialah
T = 2π L . Cari dT dan seterusnya hitung
kedalaman cecair itu. 10 dL
perubahan dalam L jika T berubah daripada
Tunjukkan bahawa perubahan hampir dalam
0.8 s kepada 0.825 s.
isi padu cecair apabila kedalaman cecair
The period of oscillation, T s, of a pendulum with a
berubah daripada 10 cm kepada 10.01 cm
length of L cm is given by T = 2π L . Find dT
ialah 1.16 cm3 10 dL

The volume of a container, V cm3, is given by and then calculate the change in L if T changes from
V = x(6x − 4), where x cm is the depth of the liquid.
0.8 s to 0.825 s.
Show that the approximate change in the volume of
the liquid when the height changes from 10 cm to 1
10.01 cm is 1.16 cm3.
L2
V = 6x2 – 4x T = 2π L = 2π
10
dV 10
dx dT 1
= 12x –4 dL = π L δT = 0.025
10
dV dT
δV = dx · δx δx = 0.01, x = 10 δT = dL . δL

= (12x – 4)δx 0.025 = π δL
= (120 – 4)(0.01) 10(0.8)
= 1.16 cm3
0.025 8 = δL
π

δL = 0.0225 cm

51

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan SPM 12

PRAKTIS

BAB 2 Kertas 1 Pada P(3, 0), dy = 4 = 1
dx (3 + 1)2 4
dy At
1. Diberi x = 3t3 − 2 dan dt = 15t2, cari
q = 1
dy 4
2014 Given x = 3t3 − 2 and dt = 15t2, find

(a) dx dalam sebutan t.
dt

dx in terms of t.
dt

(b) dy 3. Rajah menunjukkan suatu graf kubik y = f(x) dan
dx graf linear y = g(x). Titik-titik A, B, C dan D terletak

x = 3t3 – 2, dy = 15t2 2017 pada lengkung manakala E terletak pada garis
dt lurus. Tangen pada B dan D adalah selari dengan
paksi-x.
(a) dx = 9t2
dt The diagram shows a cubic graph y = f(x) and a linear
graph y = g(x). The points A, B, C and D lie on the curve
(b) dy = dy · dt = 15t2 × 1 while E is on the line. The tangents at B and D are parallel
dx dt dx 9t2 to the x-axis.
5
= 3 y

2. Rajah menunjukkan sebahagian daripada A CD
x – 3 B
2016 lengkung y = x + 1 dan satu garis lurus y = qx + 2. x
OE
x–3
The diagram shows a part of the curve y = and the
straight line y = qx + 2. x+1

y Nyatakan titik yang manakah memenuhi syarat-

y = qx+2 x–3 syarat berikut:
x+1
y = State the point(s) which satisfy the following conditions:

P (a) dy = 0
O dx
x
dy
(b) dx 0

(c) dy 0
dx
Diberi bahawa garis lurus itu selari dengan tangen
( a) ddyx = 0 pada titik B dan D.
kepada titik P pada lengkung itu. Cari nilai q. ( b) ddyx
at point B and D.
Given that the straight line is parallel to the tangent at
point P on the curve. Find the value of q.

y = x –3 ( c) ddyx  0 pada titik C dan E.
x +1
at point C and E.

dy = (x + 1) – (x – 3) = (x 4  0 pada titik A.
dx (x + 1)2 + 1)2
at point A.

Apabila y = 0, x = 3 P(3, 0)

When

52

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan 

4. Carikan koordinat titik pegun bagi lengkung (b) persamaan garis normal

y = x(x − 2)2 dan tentukan sifat bagi titik-titik itu. Equation of the normal line

Find the coordinates of the stationary points of the curve y = – 5 x + c
y = x(x − 2)2 and determine the characteristics of these 2
points. BAB 2
0 = – 5 (2) + c
2
y = x(x – 2)2
ddyx = 2x(x –2) + (x – 2)2 c = 5
= (x – 2)(2x + x – 2) = (x – 2)(3x – 2)
∴ y = – 5 x + 5
2

dy = (x – 2)(3x – 2) = 0
dx
2
x = 2 ; x= 3 2. Rajah menunjukkan sebuah rantau tertutup yang
dibina daripada seutas dawai dengan panjang
y = 0 2  2 – 22
3 3 2016 120 cm.

= 32 The diagram shows an enclosed area that is built by a
27 wire of the length 120 cm.
d2y
dx2 = (x – 2)(3)+ (3x – 2) AF

= 6x – 8 BE

Apabila x = 2 d2y = 12 – 8 = 4  0
dx2
When CD

∴(2, 0) ialah titik minimum Diberi AF = CD dan ABC dan DEF adalah dua
semibulatan dengan jejari j cm.
is a minimum point
Given that AF = CD and ABC and DEF are two semicircles
 Apabila x = 2 , d2y = 6 2 –8 with radius j cm.
3 dx2 3
When (a) Tunjukkan bahawa luas rantau tertutup, L
ialah L = (120j − 2πj2) cm2
= –4  0
Show that the enclosed area L is L = (120j − 2πj2) cm2
  2, 32 ialah titik maksimum
3 27 (b) Cari nilai j supaya luas rantau tertutup L ialah
is a minimum point maksimum. Cari nilai L tersebut.

Kertas 2 Find the value of j so that the enclosed area L is
maximum. Hence find the value of L.
2x – 4
1. Diberi bahawa persamaan lengkung y = 1 + x2 (a) Perimeter = 120 = 2πj + 2x
dan lengkung ini melalui titik P(2, 0).
2015 60 = πj + x
2x – 4 A x F
Given the equation of the curve y = 1 + x2 and the C E j
Luas, L = x(2j)

curve passes through the point P(2, 0). Area = 2j[60 – πj] B

(a) Cari kecerunan pada titik P. = 120j – 2πj2 xD

Find the gradient at the point P. (b) dL = 120 – 4πj = 0
dj
(b) Hitung persamaan garis normal kepada

lengkung pada titik P. j = 120 = 30 cm
Calculate the equation of the normal line to the 4π π

curve at point P. d2L
dj2
dy 2(1 + x2) – (2x – 4)(2x) = –4π  0
dx (1 + x2)2
(a) = ∴ Luas ialah maksimum

= 2[1 + x2 – 2x2 + 4x] The area is maximum
(1 + x2)2
Apabila j = 30 cm
dy 2[1 + 4 – 8 + 8] π Praktis
Apabila x = 2, dx = 25 SPM
L = 120 30  – 2π 30 2 = 1800 Ekstra
dy 2 π π π
dx = 5

53

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan KBAT KBAT

Sudut Ekstra

BAB 2 1. Air mengalir keluar daripada sebuah kon yang 2. Rajah menunjukkan lengkung y = x2 dan titik

tertonggeng dengan kadar 6 cm3 s−1. Jejari tapak R(−1, 0). Titik P(p, 0) adalah satu titik yang

dan tinggi kon masing-masing ialah 10 cm dan berubah pada paksi-x dan p bertambah dengan

15 cm. Hitung kadar 10 unit s−1. PQ adalah selari dengan paksi-y.
The diagram shows a curve y = x2 and point
Water flows out of an inverted cone at a rate of R(−1, 0). Point P(p, 0) is a variable point on the x-axis
6 cm3 s−1. The radius and the height of the cone are and p increases at a rate of 10 unit s−1. PQ is parallel
10 cm and 15 cm respectively. Calculate
to the y-axis. y
(a) kadar perubahan ketinggian paras air yang
y = x2
berkurang Q

the rate of change of the height of the water that x
is decreases
R(–1, 0) O P(p, 0)
(b) kadar perubahan luas permukaan air yang

berkurang

the rate of change of the surface area of the water
that is decreases

apabila paras air adalah setinggi 9 cm.

when the height of the water is 9 cm.

(a) dV = –6 cm3 s–1 (a) Nyatakan luas, A unit2, bagi PQR dalam
dt sebutan p.
10 cm
r = h r= 2 h State the area, of A unit2, for PQR in terms of p.
10 15 3 r
h 15 cm (b) Hitung kadar perubahan A yang bertambah
V = 1 πr 2h pada ketika p = 2.
3
Calculate the rate of change of A that is increasing
= 1 π  2 h2h when p = 2.
3 3
(c) Hitung kadar perubahan bagi panjang PQ
V = 4 π h3 pada ketika p = 1.
27
Calculate the rate of change of the length of PQ
dV = 4 π h2 when p = 1.
dh 9
(a) dp = 10 unit s–1 y
dV = 4 π h2· dh dt y = x2
dt 9 dt 1
A= 2 [p + 1]p2
4 dh
–6 = 9 π [9]2 dt A= 1 p3 + 1 p2 p2
2 2 P(p, 0)
–6 = dh dh =– 1 cm s–1 dA 3 R(–1, 0) 0 x
36π dt dt 6π dp 2
(b) = p2 + p

(b) L = πr2 = π[ 2 h]2 dA =  3 p + p · dp
L = 3 dt 2 dt
4
9 πh2   dA = 3 + = unit2
p = 2, dt 2 (2)2 2 10 80 s–1
ddhL = 8
9 πh (c) p2 = PQ d(PQ) dp
dt dt
ddLt = 8 πh dh d(PQ) = 2p = 2p ·
9 dt dp
p = 1, d(PQ) = 2(1)(10)
  = 8 π[9] – 1 dt
9 6π
= 20 unit s–1
= – 4 s–1
3 cm2

Quiz 2

54

BAB   Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran 

3NOTA IMBASPAeNngamiran BAB 3
Integration

3.1 Pengamiran Sebagai Songsangan Pembezaan

Integration as the Inverse of Differentiation

NOTA IMBASAN

1. Dalam proses pembezaan, jika y = 1 x2 + c, dengan c 2. Proses ini adalah songsangan bagi pembezaan dan
2 dikenali sebagai pengamiran.
dy
ialah pemalar, maka fungsi kecerunan dx = x. Secara This process is the inverse of differentiation and is called
integration.
dy
songsang, jika diberikan fungsi kecerunan dx = x, 3. Tata tanda bagi pengamiran ialah ∫ f(x)dx.

maka persamaan lengkung ialah y = 1 x2 + c. ∫ The notation for integration is f(x) dx.
2 dy
Sebagai contoh, dx = x
In the process of differentiation, if y = 1 x2 + c, where c is a For example,
2
dy = x dx
constant, then the gradient function is dy = x. Conversely, if the
dy dx ∫dy = ∫ x dx

gradient function is given by dx = x, then the equation of the y = 12 x2 + c.
1
curve is y = 2 x2 + c.

1. Cari kamiran melalui songsangan pembezaan terhadap x.

Find the integral by inversing the differentiation with respect to x. TP 2

CONTOH 1 (a) Diberi/Given d ( 2 x6) = 4x5, (b) Diberi/Given dy 4f(x) = g(x),
dx 3 dx

Diberi/Given d (3x3) = 9x2, cari/find ∫ 4x5 dx. cari/find 3∫ g(x) dx.
dx

∫Cari/Find 9x2 dx.

Penyelesaian: ddx  2 x6 = 4x5 3∫g(x)dx = 3[4f(x)] + c
3
d (3x 3) = 9x 2 = 12f(x) + c
dx
d 2 x6 = 4x5·dx
d(3x3) = 9x2·dx 3

∫d(3x3) = ∫ 9x2 dx ∫d 2 x6 = ∫4x5 dx
3

∫9x2 dx = 3x3 + c ∫4x5 dx = 2 x6 + c
3

Tip

∫ ∫9x2dx boleh ditulis sebagai 9 x2dx.
∫ ∫9x2dx can be written as 9 x2 dx.

55

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran

CONTOH 2 (c) Diberi d (2x – 5) = 1 h(x), (d) Diberi y = (x − 1)(x − 3)2,
dx (3 – 4x)3 2
dy cari dy . Seterusnya, cari
Diberi y = x2(2 − x)2, cari dx . cari 1 ℎ(x)dx, dengan a ialah dx
∫ a
BAB 3 ∫(x − 3)(3x − 5) dx.
Seterusnya, cari pemalar.
Given y = (x − 1)(x − 3)2, find ddyx .
∫ x(2 − x)(4 − 3x)dx. Given d (2x – 5) = 1 h(x),
dx (3 – 4x)3 2 Then, find ∫(x − 3)(3x − 5)dx.
dy
Given y = x2(2 − x)2, find dx . find ∫ 1 ℎ(x)dx, such that a is a
a y = (x − 1)(x − 3)2
Then, find ∫ x(2 − x)(4 − 3x) dx. constant.
ddyx = 2(x − 1)(x − 3) + (x − 3)2
Penyelesaian: ∫ 1 h(x)dx = (2x – 5) +c = (x − 3)(2x − 2 + x − 3]
y = x2(2 − x)2 2 (3 – 4x)3 = (x − 3)(3x – 5)

dy = −x2(2 − x) + 2x(2 − x)2 ∫(x − 3)(3x − 5)dx
dx
1 2  (2x – 5)  = (x − 1)(x − 3)2
= x(2 − x)[−x + 2(2 − x)] ∫ a h(x)dx = a (3 – 4x)3 + c

= x(2 − x)(4 −3x)

Maka/Hence

∫ x(2 − x)(4 − 3x) dx = x2(2 − x)2

CONTOH 3 (e) Diberi y = x 2x , cari dy . (f) Diberi y = x + 1 dan dy =f(x),
+1 dx x2 dx

Diberi y = x x , cari dy . Seterusnya, cari ∫ (x 6 1)2 dx. dengan π ialah pemalar.
2– dx +
Cari ∫ 2 f(x) dx.
Seterusnya, cari 1 dx. Given y = 2x , find dy . π
∫ (2 – x)2 x+1 dx

x dy Then , find 6 dx. Given y = x+1 and dy = f(x),
Given y = – , find dx . Then, ∫ + 1)2 x2 dx
2 x (x
such that π is a constant.
1
∫find (2 – x)2 dx. 2
π
Penyelesaian: y = 2x Find ∫ f(x) dx.
x+1
y= x
2–x = 2(x + 1) – 2x ∫f(x)dx = x+1 +c
(x + 1)2 x2
dy = (2 – x)(1) + x
dx (2 – x)2 = 2 2π ∫ f(x)dx = 2  x + 1  + c
(x + 1)2 π x2
= 2 1
(2 – x)2 ∫2 (2 – x)2 dx ∫ 2 dx = 2x + c
(x + 1)2 (x + 1)
∫= (2 2 x)2 dx
– 6 2
+ 1)2 dx = 3∫ + dx
∫ 2 dx = x ∫ (x (x 1)2
(2 – x)2 2–x
3(2x)
∫ 1 dx = x = (x + 1) + c
– x)2 2(2 –
(2 x) 6x
(x + 1)
= + c

56

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran 

3.2 Kamiran Tak Tentu

Indefinite Integral

NOTA IMBASAN BAB 3

1. Kamiran axn terhadap x. Katakan u = ax + b, maka

Integrate axn with respect to x. Let u = ax + b, then

axndx = axn + 1 + c, n ≠ −1, a dan c ialah pemalar. du = adx = du
axndx = (n + 1) ≠ −1, where a and c are constants. dx a
∫
axn + 1 + c, n ∫ (ax + b)ndx = ∫un du = un + 1 +c
(n + 1) a a(n + 1)

∫

2. Kamiran bagi ungkapan algebra = (ax + b)n + 1 n ≠ −1,
Integration for algebraic expressions a(n + 1)
,
∫[f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx
a, b dan n ialah pemalar.
3. Kamiran (ax + b)n terhadap x. a, b, and n are constants.
Integrate (ax + b)n with respect to x.

2. Cari kamiran yang berikut terhadap x.

Find the integral with respect to x of the following. TP 2

CONTOH (a) ∫ 3 x dx
4

(i) ∫ 1 dx (ii) ∫ x dx ∫ 3 x dx = 3 x2 +c
2 4 4 2

Penyelesaian: = 3 x2 + c
8
(i) ∫ 1 dx = 1x 0 + 1 +c= 1 x + c
2 2 2

1 mesti tukar kepada
bentuk axn.
(ii) ∫ x dx = x 2 dx must convert to axn.
x1
2 +1

= 1 +c
2
+1 Tip

= 2 x x +c x 1 + 1 = (x)(x) 1
3 2 2

= x x

(b) ∫ 2 dx (c) ∫ x dx (d) ∫ 5x dx
3x3 (2x)4 2x2

∫ 2 dx = ∫ 2 x–3 dx ∫ x dx = ∫ 1 x –3 dx 1 3
3x3 3 (16x4) 16
∫ 5x 2 dx = ∫ 5 x– 2 dx
2 x –2 1 1 2x2 2
= 3 –2 +c = 16 (–2)x2 +c
= 5 × (–2) +c
2 x
= – 31x2 + c = – 321x2 + c
= – 5x + c

57

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran

(e) ∫ x dx (f) ∫ (2x)3 dx (g) 2 ∫ x2 dx
3x 4x5 3 x

BAB 3 ∫ x dx = ∫ 1 x 1 ∫ 8x3 dx = ∫2x–2dx 2 ∫ x2 dx = 2 3
3x 3 4x5 3 3
2 dx –2 1 ∫x 2 dx
x
= 1 × 2 x x +c = +c x2 = 2 × 2 x2 x +c
3 3 3 5


= 2 x x +c = 4 x2 x +c
9 15

3. Tentukan kamiran tak tentu bagi setiap fungsi algebra yang berikut.

Determine the indefinite integral for each of the following algebraic expression. TP 2

CONTOH (a) ∫(2x2 – 4x + 1 )dx
x2
+ 1 dx (ii) ∫ 3x(x52 – 4) dx
(i) ∫ x 4

∫2x2dx – ∫4xdx + ∫x–2dx
Penyelesaian: 2 1
= 3 x3 – 2x2 – x + c
x +1 x 1
(i) ∫ 4 dx = ∫ 4 dx + ∫ 4 dx (ii) ∫ 3x(x2 – 4) dx
5
1 x2 1
= 4  2  + 4 x + c = ∫ 3x3 dx – ∫ 12x dx
5 5
1 1 3x4 12x2
= 8 x2 + 4 x + c = 5(4) – 5(2) +c

= 3x4 – 6x2 +c
20 5

(b) ∫ 5(x2 – 2 ) dx (c) ∫ 3x2 – 4 dx (d) ∫  2 –6+ 3  dt
x2 3 x2 t3 t2

∫5x2dx – ∫ 10 dx ∫ 3x2 dx – ∫ 4 dx = ∫2t–3dt – ∫6 dt + ∫3t –2dt
x2
2 2 = –t–2 – 6t – 3t–1

= 5 x3 + 10 + c x3 x3 = – t12 – 6t – 3 + c
3 x 4 – 2 t

= ∫3x 3 dx – ∫4x 3 dx
3 7 1
= 7 3x 3  –
34x 3 

= 9 x 7 – 1 +c
7 3
12x 3

(e) ∫(2x + 3)(4 – 2x)dx (f) ∫ 8x(x2 – 9) dx (g) ∫ x(x2 – 4x – 5) dx
5(x + 3) (x – 5)

∫(8x – 4x2 + 12 – 6x) dx 8 ∫(x2 – 3x)dx ∫ x(x – 5)(x + 1) dx
5 (x – 5)
= ∫(2x – 4x2 + 12) dx
4 = 8  x3 – 3x2  + c = ∫(x2 + x)dx
= x2 – 3 x3 + 12x + c 5 3 2

= 1 x3 + x2 +c
3 2

58

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran 

4. Tentukan kamiran tak tentu bagi fungsi berikut yang berbentuk (ax + b)n, dengan keadaan a dan b ialah BAB 3
pemalar, n ialah integer dan n ≠ –1.

Determine the indefinite integral for the following functions in the form of (ax + b)n, where a and b are constants, n is an

integer and n ≠ –1. TP 3

CONTOH (a) ∫ 3 (x – 5)3dx (b) ∫ (3 2 dx
4 – 4x)2

∫ (2x + 3)4 dx

Penyelesaian: Katakan u = x – 5 Katakan u = 3 – 4x

Katakan/Let u = 2x + 3 du = 1, dx = du du = –4, dx = du
dx 1 dx –4
du = 2, dx = du
dx 2 = 3 u3 du = ∫2u–2 du = u– 2 + 1 +c
∫ 4 1 –4 –2(–1)
= u4 du = u4 + 1 + c
∫ 2 2(4 + 1) 3(x – 5)4 1
= 16 +c = – + c
(2x + 3)5 2(3 4x)
= 10 +c

(c) ∫ 6 x – 2 dx (d) ∫ 2 x)3 dx (e) ∫ 3 dx
3(3 – 1 – 2x

Katakan u = x – 2 Katakan u = 3 – x Katakan u = 1 – 2x

du = 1, dx = du du = –1, dx = –du du = –2, dx = du
dx dx dx –2
3
1 –2 –2u–3 + 1 – 1 + 1
du = 12u 2 +c = ∫ 3 u–3du = 3(–2) + c 3 –1
= ∫6u 2 3 –2 3u 2
= ∫ u 2 du = +c

= 4(x – 2) x – 2 + c = 1 +c –2 1 
3(3 – x)2 2
= –3 1 – 2x + c

5. Tentukan persamaan lengkung, y daripada fungsi kecerunan yang berikut.

Determine the equation of the curve, y from the following gradient functions. TP 3

CONTOH (a) Diberi dy = 3 − x2 – 4x dan y = 4 apabila x = 0.
dx
dy
Diberi dx = 3x2 – 5 dan y = 1 apabila x = −1. Given that dy = 3 − x2 – 4x and y = 4 when x = 0.
dx
Given that dy = 3x2 – 5 and y = 1 when x = −1.
dx dy = 3 – x2 – 4x
Penyelesaian: dx
ddyx = 3x2 − 5 Persamaan lengkung ialah
fungsi y dalam sebutan x ∫dy = ∫(3 – x2 – 4x)dx
Gunakan x dan y untuk 1
mendapat nilai c. 3
The equation of the curve is
∫dy = ∫(3x2 − 5)dx function y in terms of x. Use x y = 3x – x3 – 2x2 + c

y = x3 − 5x + c and y to obtain a value of c. Apabila x = 0, y = 4

Apabila/When x = −1, y = 1 c=4

1 = −1 + 5 + c Persamaan lengkung ialah

c = −3 The equation of the curve is

Persamaan lengkung ialah y = x3 − 5x – 3. y = 3x – 1 x3 – 2x2 + 4
The equation of the curve is y = x3 − 5x – 3. 3

59

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran

(b) Diberi dy = x – 2 dan y = 1 apabila x = 6. (c) Diberi dy = x2(2x + 1) dan graf melalui titik
dx dx

Given that dy = x – 2 and y = 1 when x = 6. (1, −1).
dx
BAB 3 Given that dy = x2(2x + 1) and the graph passes
dx

through the point (1, −1).

dy 1 dy = 2x3 + x2
dx dx
= (x – 2) 2
1
∫dy = ∫(2x3 + x2)dx
∫dy = ∫(x – 2) 2 dx 1 1
3 y = 2 x4 + 3 x3 + c
2
y = 3 (x – 2) 2 + c Apabila x = 1, y = –1

Apabila x = 6, y = 1 –1 = 1 + 1 +c
2 3
1 = 16 + c
3 11
c = – 6
13
c = – 3 Persamaan lengkung ialah

Persamaan lengkung ialah The equation of the curve is

The equation of the curve is y= 1 x4 + 1 x3 – 11
2 3 6
y= 2 (x – 3 – 13
3 3
2) 2

(d) Diberi dy = (ax − 3) dan graf melalui titik (e) Diberi dy = 3x2 + b dan graf melalui titik (1, 3)
dx dx

(−1, 8) dan (3, 4), dengan a ialah pemalar. dan (−1, −3), dengan b ialah pemalar.

Given that dy = (ax − 3) and the graph passes Given that dy = 3x2 + b and the graph passes
dx dx
through the points of (−1, 8) and (3, 4), where a is a
through the points (1, 3) and (−1, −3), where b is a
constant.
constant.

dy = ax – 3 dy = 3x2 + b
dx dx

∫dy = ∫(ax – 3)dx ∫dy = ∫(3x2 + b)dx
a
y = 2 x2 – 3x + c y = x3 + bx+ c

Apabila x = –1, y = 8 Apabila x = 1, y = 3

8= 1 a + 3 + c 3 = 1 + b + c
2
2 = b + c ...➀

5= 1 a + c ...➀ Apabila x = –1, y = –3
2
–3 = –1 – b + c

Apabila x = 3, y = 4 –2 = –b + c ...➁

4= 9 a – 9 + c ➁ – ➀: –4 = –2b
2
b = 2, c = 0

13 = 9 a + c ...➁ Persamaan lengkung ialah
2
The equation of the curve is

➁ – ➀: 8 = 4a y = x3 + 2x

a = 2, c = 4

Persamaan lengkung ialah

The equation of the curve is

y = x2 – 3x + 4

60

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran 

3.3 Kamiran Tentu

Definite Integral

NOTA IMBASAN BAB 3

1. Nilai ∫ f(x) dx dapat ditentukan apabila x = a digantikan 5. (i) Luas dibatasi oleh lengkung y = f(x) dan satu graf
ke dalam persamaan ∫f(x)dx = F(x) + c, iaitu apabila lagi y = g(x) dari x = a ke x = b diberi oleh:
x = a, ∫ f(x)dx = F(a) + c.
Area of the region bounded by the curve y = f(x) and another
The value of ∫ f(x)dx can be obtained when x = a is graph y = g(x) from x = a to x = b is given by:
substituted into the equation ∫f(x) dx = F(x) + c, that is when
x = a, ∫ f(x)dx = F(a) + c. yyy

2. Pengamiran fungsi f(x) terhadap x dari x = a hingga y = f(x) y = g(x)
x = b ialah ditulis seperti: =–

The integration of the function f(x) with respect to x from x = a to Oa b x Oa b x Oa bx
x = b is written as:

∫abf(x) dx = [F(x) + c]ba = [F(b) + c] – [F(a) + c] = F(b) – F(a).

3. Beberapa petua bagi kamiran tentu. (ii) Luas dibatasi oleh lengkung x = f(y) dan
satu graf lagi x = g(y) dari y = a ke y = b.
Rules for definite integral
Area of the region between the curve x = f(y) and another
(i) ∫ab[f(x) ± g(x)] dx = ∫abf(x) dx ± ∫abg(x)dx curve x = g(y) from y = a to y = b is given by:
d
(ii) Jika/If dx [f(x)] = f(x), maka/then ∫abf(x)dx = F(b) – F(a)

(iii) ∫aaf(x) dx = 0 y yy

(iv) ∫ca f(x) dx + ∫bcf(x) dx = ∫ b f(x) dx b b x = f(y) b x = g(y)
a

(v) N∫Oba fT(xA) dIxM=B–∫Aab fS(xA) dNx = dy – dy

4. Luas di bawah suatu lengkung a aa
O xO xO x
Area under the curve

(i) luas rantau di antara lengkung y = f(x) dan paksi –x

dari x = a ke x = b diberi oleh:

Area of the region between the curve y = f(x) and the x-axis 6. Isi padu janaan

from x = a to x = b is given by: Generated Volume

∫ba y dx = ∫ba f(x)dx (i) Apabila kawasan berlorek dalam rajah diputarkan

(ii) luas rantau di antara lengkung x = g(y) dan paksi-y melalui 360° pada paksi-x, isi padu janaan ialah

dari y = a ke y = b diberi oleh: I = ∫πb y2 dx.
a
Area of the region between the curve x = g(y) and the y-axis When the shaded region in the diagram is rotated through

from y = a to y = b is given by: ∫360° about the x-axis, the generated volume is I = π b y2 dx.
a
b b
A = ∫ a x dy = ∫ a g(y) dy

yy y y = f(x)

y = f(x) x = g(y) dx
y
b
dy x x

y a Oa b
x
Oa dx b x O x

(i) (ii)

61

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran

(ii) Apabila kawasan berlorek dalam rajah diputarkan y

melalui 360° pada paksi-y, isi padu janaan ialah
b
I = ∫π a x2 dy. b x = f(y)

BAB 3 When the shaded region in the diagram is rotated through
b
∫360° about the y-axis, the generated volume is I = πa x2 dy. x dy
a
x
O

6. Nilaikan yang berikut.

Evaluate the following. TP 3

CONTOH (a) ∫ 0x (2x – 1)dx (b) 3 4x3 – 3 dx
x2
–2 ∫1

∫3–1(x – 1)(x + 3)dx ∫0 x(2x – 1)dx = ∫0–2(2x2 – x)dx ∫3 4x3 – 3 dx = ∫13(4x – 3x–2)dx
–2 1 x2
Penyelesaian:
 =2 x3 – 1 x2 0  = 3 3
∫3–1(x – 1)(x + 3)dx 3 2 –2 2x 2 + x 1

= ∫3–1(x2 + 2x – 3)dx = 0 – –136 – 2 = (18 + 1) – (2 + 3)

 = 1 3 22 = 14
3 –1 3
x3 + x2 – 3x =

= [9 + 9 – 9] –  – 13 + 1 + 3

= 5 13

 (c) 1 x + 2 dx (d) 4 (2 2 x)3 dx (e) 2 4x2 – 9 dx
x3 –
∫ –2 ∫0 ∫–2(2x – 3)

∫1–2(x + 2x–3)dx ∫ 4 2(2 – x)–3dx 2 (2x + 3)(2x – 3) dx
0 (2x – 3)
∫–2
 = 1 1 2(2 – x)–2 4
2 x2 (–2)(–1) 0
x 2 – 1  = = [x2 + 3x]–22
–2

=  1 – 1 – 2 – 1  = 1 – 1 =0 = (4 + 6) – (4 – 6)
2 4 22 22
= 12

= 3 –3
4

= –2 14

62

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran 

7. Cari nilai bagi yang berikut.

Find the value of the following. TP 4

CONTOH BAB 3

Diberi 4 f(x)dx = 4, 4 f(x)dx = 2 dan 4 g(x)dx = 5, cari nilai bagi

∫–2 4 ∫0 4 ∫–2 4
–2 0 –2
f(x) dx = 4, f(x) dx = 2 and g(x) dx = 5, find the value of
∫ ∫ ∫It is given that
(i) ∫–422f(x)dx 2 ∫4
(iv) ∫ –2 f(x)dx + 2 f(x)dx

(ii) ∫ [4 f(x) – g(x)]dx (v) k jika/if ∫–42[k − f(x)]dx = 8
–2

(iii) ∫–02 f(x)dx

Penyelesaian:

(i) ∫–422f(x)dx = −8 (iv) ∫ 2 f(x)dx + ∫4 f(x)dx
–2 2

(ii) ∫4 [ f(x)dx – g(x)]dx = ∫4 f(x)dx = 4
–2 –2

= ∫4 f(x) dx − ∫4 g(x)dx (v) ∫–42[k − f(x)]dx = 8
–2 –2
= 4 – (5) = −1
∫–42kdx – ∫4 f(x)dx = 8
(iii) ∫ 0 f(x)dx –2
–2
[kx]−42 = 8 + 4
= ∫4 f(x)dx – 4 f(x)dx 4k + 2k = 12
–2
=4–2=2 ∫0

k = 2

(a) Diberi 3 f(x) = 6, –1 f(x)dx = −1 dan –1 g(x)dx = 4, cari nilai bagi

∫ –1 ∫ –3 ∫ –3

Given that 3 f(x) = 6, –1 f(x)dx = −1 and –1 = 4, of
∫ ∫ ∫ –1 –3 –3 g(x)dx find the value

(i) ∫ 3 f(x) dx (iv) ∫ 2 1 f(x)dx + 3 1 f(x)dx
–3 –1 2 2
∫2

(ii) ∫ –1 [g(x) – 2f(x)]dx (v) k jika/if ∫––31[2g(x) − kx]dx = 20
–3

(iii) ∫ 0 f(x) + 3 f(x)dx
–3
∫0

(i) ∫ –33f(x)dx = –1 f(x)dx + 3 f(x)dx  (iv) 12 2 + 3
∫ –3 ∫ –1 f(x)dx ∫ 2 f(x)dx
∫ –1

= –1 + 6 = 5 = 1 [6] = 3
2
(ii) ∫ –1 g(x)dx – 2∫ –1 f(x)dx
–3 –3   (v ) ∫2––(3142)g–(x)kd2xx2∫– –1 kx dx = 20
–3 20
= 4 – 2(–1) –1 =
–3
=6

( iii) ∫ 0 f(x)dx + ∫ 3 f(x)dx 8 –  k – 9k  = 20
–3 0 2 2

= ∫ 3 f(x)dx = –1 + 6 = 5 4k = 12
–3

k = 3

63

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran

8. Selesaikan yang berikut.

Solve the following. TP 4

BAB 3 CONTOH (a) Diberi ∫–k2(x − 4)dx = −10, cari nilai k, dengan

Diberi ∫ k1(3x − 2)dx = 5 , cari nilai k, dengan keadaan k ≥ 0.
2
keadaan k > 0. ∫ Given that k–2(x − 4)dx = −10, find the value of k,

where k ≥ 0.

∫Given that k (3x − 2)dx = 5 , find the value of k, where ∫k–2(x – 4)dx = –10
1 2

k > 0.   x2 k =
2
Penyelesaian: – 4x –2 –10

∫ k1(3x − 2)dx = 5  k2 – 4k – (2 + 8) = –10
2 2

 3x2 – 2x k1 = 5 k(k – 8) = 0
2
2 k = 0 atau k = 8

32k2 – 2k –  3 – 2 = 5
2 2

3k2 – 4k + 1 = 5

3k2 – 4k – 4 = 0

(3k + 2)(k – 2) = 0

k = 2

(b) Diberi ∫ 2k( 2 )dx = 3 , cari nilai k. (c) Diberi d  (x 1 4 = g(x), cari nilai bagi
x3 4 dx – 2)2

∫ 2k( 2 )dx 3 find the k. ∫4–1[2x – g(x)]dx.
Given that x3 = 4 , value of

d 1
dx (x – 2)2
3  = g(x), find the value of
Given that

∫4–1[2x – g(x)]dx.

2 2x–3dx = 3 ∫ –412xdx – 4 g(x)dx
4
∫k ∫–1

14
(x – 2)2 –1
  – x12 2 = 3  = [x2]4–1 –
4
k 3 1 41
4
 – 14  –  – k12  = 3 =(16 – 1) – – 9 –1
4
5 1 4 3361
1 – 1 = 3 = 15 – 36 =
k2 4 4

4 – k2 = 3k2

4k2 = 4

k = ±1

64

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran 

(d) Diberi d2y = 1 − 6x2 apabila dy = −1, x = −1 (e) Diberi d2y = 4x2 − 2 apabila dy = 4, x = 0 dan
dx2 dx dx2 dx
dan y = 2, cari nilai y dalam sebutan x. y = 1, cari nilai y dalam sebutan x.

Given that d2y =1 − 6x2 when dy = −1, x = −1 and Given that d2y = 4x2 −2 when dy = 4, x = 0 and BAB 3
dx2 dx dx2 dx
y = 1, find the value of y in terms of x.
y = 2, find the value of y in terms of x.

d2y = 1 – 6x2 dy = x – 2x3 – 2 dy = 4 x3 – 2x + c
dx2 dx dx 3

dy = x – 2x3 + c y= x2 – 1 x4 – 2x + c, x = 0, dy = 4, c = 4
dx 2 2 dx
1
x = –1, dy = –1 x = –1 y = 2 y= 3 x4 – x2 + 4x + c
dx
2= 1 – 1 + 2 + c, x = 0, y = 1, c1 = 1
2 2
–1 + 2 + c = –1 1
c = –2 c1 = 0 1 1 ∴y = 3 x4 – x2 + 4x + 1
2 2
∴y = x2 – x4 – 2x

9. Cari kamiran tentu atau luas berdasarkan rajah yang diberi.

Find the definite integral or area based on the diagram given. TP 3

CONTOH (a) y

y

y = f(x) –2 O P y = f(x) x
A Q

37

aO bBc x

Diberi luas P ialah 4.5 unit2 dan 7 f(x)dx =

Diberi luas rantau A ialah 10 unit2 dan luas B 8 unit2, cari ∫ −2

Given that the area of P is 4.5 unit2 and 7
8 unit2, find −2
ialah 6 unit2, cari ∫ f(x)dx =

Given that the area of the region A is 10 unit2 and the (i) ∫ 7 f(x)dx
area of B is 6 unit2, find 3

(i) ∫ c f(x)dx (ii) ∫ 3 f(x)dx
b −2

(ii) ∫ c f(x)dx (iii) jumlah luas P dan Q.
a
the total area of P and Q.
| |(iii) ∫b + c
a f(x)dx f(x)dx
∫b

Penyelesaian:
c
( i) ∫ b f(x)dx = Luas di bawah paksi-x (i) 7 f(x)dx = 8 + 4.5

= Area under the x-axis ∫3 = 12.5 unit2

=6 b ∫c
∫c a b
(ii) a f(x)dx = ∫ f(x)dx + f(x)dx (ii) 3 f(x)dx = −4.5 unit2

= 10 – 6 = 4 ∫ –2

| |(iii) ∫b + ∫c Nilai ialah –6.
a f(x)dx b f(x)dx Value is –6. (iii) Jumlah luas = 12.5 + 4.5
Total area = 17 unit2
= 10 + |–6| = 16

65

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran

(b) y (c) y

3 5 x = f(y)
H Q
BAB 3
K x 1P
y = f(x)
–4 O 2 x
5
O

Diberi 0 f(x)dx = 6 dan 2 f(x)dx = 5. Diberi luas Q ialah 18 unit2,

∫ –4 0 ∫0 2 Given that the area of Q is 18 unit2,
–4 0
f(x)dx = 6 and = 5. (i) cari luas P,
∫ ∫Given that f(x)dx
find the area of P,
Cari / Find
(ii) ungkapkan luas Q dalam tatatanda
(i) luas H / the area of H
kamiran.
(ii) luas K / the area of K
express the area of Q in integration notation.
(iii) ∫ 2 f(x)dx
–4

(i) Luas H = luas segiempat (i) Luas P/ Area P = 25 − Q
Area H = Area of rectangle
0 = 25 − 18
∫ – –4 f(x)dx
= 7 unit2
= 12 – 6 = 6 unit2
(ii) Luas Q = ∫ 5 x dy
(ii) Luas K = 2 f(x)dx = 5 1
Area Q
∫0
2 0 2
(iii) ∫ –4 f(x)dx = f(x)dx + f(x)dx
∫ –4 ∫0
= 6 + 5 = 11 unit2

10. Cari luas rantau berlorek bagi setiap rajah yang berikut.

Find the area of the shaded region for each of the following diagrams. TP 4

CONTOH y (a) y = 2 + x – x2 y
y = 2 + x – x2
y = x2 – 2x y = x2 – 2x
A

Penyelesaian Guna/Use
y = x2 – 2x
= x[x – 2] ∫b ydx
a

luas berlorek/shaded area O x O Bx
3
| |=∫ 2 y dx + 3 y dx
0 Apabila x = 0 y = 2
∫2 y=0
| |= ∫20(x2 – 2x)dx + ∫32(x2 – 2x)dx 2 + x – x2 = 0
x3 x3 (2 – x)(1 + x) = 0
|  |  =3 – x2 2 + 3 – x2 3 x = 2; –1
0 2 B(2, 0) A(0, 2)

= |1 8 – 42| + Tip
3
Perlu buat kamiran berasingan kerana
1237 – 92 –
luas di bawah paksi-x ialah negatif. 2 = ∫20(2
 luas =ydx + x – x2)dx
8 ∫NJikeaed30tyo dx, kita dapat ∫0
1 3 – 42 perform the integration area 2
2x + x2 – x3 0
separately because the area under = 2 3
= |–34| + 0 + 4
3 4 x-axis is negative. = 4 + 2 – 8 4
3
= 2 32 unit2 ∫  30(x2 – 2x)dx = x3 – x2 3= 0
3
0

Luas tidak mungkin sifar. = 10 unit2
3
Area cannot be zero.

66

(b) y = 1   Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran 
x2
(c) y = x(x + 1)(2 – x)
y
y

(1, 1) y = x(x + 1)(2 – x) BAB 3
–1 O
y = 1 ΃3, 1 ΃
x2 9
x
2

Ox

alureaas==∫∫1313x1y2ddxx y = (x2 + x)(2 – x)

= –x3 + 2x2 + 2x – x2

y = –x3 + x2 + 2x

 luas = + ∫02(–x3 +

area

    =
∫ = 3 x –2dx ∫0–1(–x3 + x2 + 2x)dx x2 + 2x)dx
1 + x2 2
–x4 + 1 x3 + x2 0+ –x4 + x3
1 1 4 3 4 3 0
= – x 43 = – 3 4 – [–1] –1
1
  =– 152 + [–4 + 8 + 4]
1 2 3
=1– 3 = 3 unit2
= 5 + 8 = 3 112 unit2
12 3

(d) y = (x + 2)(3 – x) (e) y = 4 – x2 y

y –3 y = 4 – x2

y = (x + 2)(3 – x) O x
4
O1 P4 x

Apabila y = 0 Apabila y = 0

(x + 2)(3 – x) = 0 4 – x2 = 0

x = 3; –2 x = ±2

P[3,0] Luas dari x = –3 ke –2

Luas di atas paksi-x/Area above the x-axis 3 4∫= x3
––23(4 – x2)dx = 3 –2
= 3 (x + 2)(3 – x)dx = 3 (6 + x – x2)dx 4x – –3

∫1 1 4 3x3 ∫1 9 1 1
2 2 2 3
= 6x + x2 – 3 3 = 18 + – 94 – 6 + – 4 = 31–8 + 8 2 – (–12 + 9)4
3
1

= 7 31 unit2 = –2 13 unit2

Luas dari x = 3 ke x = 4 luas dari x = 2 ke x = 4

 4∫43(6 + x – x2)dx = 1 – x3 4
6x + 2 x2 3 3 3 4= ∫42(4 – x2)dx = x3 4
4x – 3 2
64 9
= 24 + 8 – 3 4 – 9 + 2  = – 10 23 unit2

= – 2 65 Jumlah luas =  –2 13  + –1 0 32 

7 31 –2 65 = 1 0 61 unit2 Total area
 ∴Jumlah luas = +
= 13 unit2
Total area

67

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran

11. Cari luas rantau berlorek bagi setiap rajah yang berikut.

Find the area of the shaded region for each of the following diagrams. TP 4

BAB 3 CONTOH (a) x = y2 – 4y

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

4 Ox
–1

1 x = y2 – 4y
Ox
luas sebelah kiri/area on the left

x = (y – 3)2 + 2 = 4 x dy

= y2 – 6y + 11 ∫0
= ∫40(y2 – 4y)dy
luas berlorek y3 64
Guna/Use =  3 – 2y2 40 =  3 – 324 –0
Shaded area
∫ 4 x dy = –1 0 23
= ∫41(y2 – 6y + 11)dy 1

3 4=y3 4
3 – 3y2 + 11y 1  4∫0–1(y2 – 4y)dy = y3 0
3 – 2y2 –1

= 1 634 – 48 + 442 – 1 1 – 3 + 112 = 0 – 1– 31 – 22 = 2 13
3

= 9 unit2 –1 0 32 + 2 31
 Jumlah luas =

Total area
= 13 unit2

(b) y = x2 (c) y = 4
x2
y
y
y = x2
4 y = 4
x2
2

1 1
Ox 2

1 O x

x = y 2 4
y
luas = 2∫ 4 x dy x2 =
1
area
4 1 2 – 1
1
= ∫2 y2 dy x= y = 2y 2

3  =2 2 3 4 ∫luas = 2 – 1 dy
3 1 1
y2 2y 2

231 2 2 (1) area 2
= 3 (8) – 3 = 4 4y 2
1

= 23134 4 28 2
3
= unit2 =4 2– 4 = 8–4 = 4
2 2 2

= 2 2 unit2

68

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran 

12. Cari luas berlorek yang dibatasi oleh lengkung, garis lurus atau satu lagi lengkung yang diberi bagi setiap yang

berikut.

Find the shaded area bounded by the curve, the straight line or another curve that is given for each of the following. TP 4

CONTOH (a) y BAB 3
y = x2 + 1
Diberi persamaan y

lengkung y = x2 − 3x + 6 y = x2 – 3x + 6

dan garis lurus y = x + 3.
Given the equation of the
curve y = x2 − 3x + 6 and 1 y = –x + 3
the straight line y = x + 3. Ox
B
Diberi persamaan lengkung y = x2 + 1 dan garis
Penyelesaian y=x+3 A C x lurus y = −x + 3.
Untuk mencari D
persilangan titik Given the equation of the curve y = x2 + 1 and the
straight line y = −x + 3.
O

x2 – 3x + 6 = x + 3

x2 – 4x + 3 = 0

(x – 3)(x – 1) = 0

x = 3 ; 1

Maka/Hence y = 6; 4 x2 + 1 = –x + 3
x2 + x – 2 = 0
A(1, 4) B(3, 6) (x – 1)(x + 2) = 0
x = 1; –2
Luas berlorek = luas ABCD – luas di bawah graf y=2
A[1, 2]
lengkung dari x = 1 ke x = 3

Luas ABCD = 3 ydx = ∫31(x + 3)dx

∫1

=  x2 + 3x 3 =  9 + 9 –  1 + 3 1
2 1 2 2
luas = ∫0 ydx + ∆ABC
= 13 21 – 3 21 = 10 unit2
area ∫1 1
0 2
= (x2 + 1)dx + (2)(2)

Luas di bawah graf lengkung 3 4 = x3 +x 1 + 2
3 0
Area under the curve

 4= ∫31(x2 – 3x + 6)dx = x3 – 3 x2 + 6x 3 = 1 1 + 12 + 2
3 2 1 3
= unit2
= 9 – 27 + 184 –  1 – 3 + 64 = 8 32 unit2 3 31
2 3 2

∴luas berlorek Kaedah Alternatif

shaded area

= 10 – 8 32 Untuk mencari luas trapezium: y
= 1 31 unit2 To find the area of trapezium: y = x2 + 1

1 × (4 + 6) × 2 A
2

= 10 unit2

Tip 1 y = –x + 3
B
• Perlu cari titik persilangan. C x
Need to find the intersection point. O dx 1 3
• Dari titik persilangan, lukis garis lurus mencancang.
From the intersection point, draw a vertical straight

line.
• Cari luas yang lebih besar di bawah garis.
Find the biggest area under the line.
• Tolak luas yang lebih kecil.
Minus the smaller area

69

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran

(b) y (c) y
y = x2
BAB 3
x = y2

Ox (–4, 0) (0, 4)
O (1, 0) x

Diberi dua persamaan lengkung, y = x2 dan Diberi persamaan lengkung y = (1 − x)(x + 4)
dan garis lurus y = x + 4.
x = y2.
Given the equation of the curve y = (1 − x)(x + 4)
Given the two equations of the curves, y = x2 and and the straight line y = x + 4.
x = y2.

Titik persilangan Titik persilangan/Intersection point

Intersection point (1 – x)(x + 4) = x + 4

x2 = x x + 4 – x2 – 4x = x + 4

x4 = x x2 + 4x = 0

x(x3 – 1) = 0 x = 0; x = 1 x(x + 4) = 0

y=0y=1 x = 0; x = –4

1 ∫1 y = 4 ; y = 0
0
 4  4luas berlorek =∫0 x dx – x2dx luas berlorek = luas segi tiga + ∫1 f(x)dx
– 0
shaded area 2 3 1
3 3 x3 1 ∫shaded area = area of triangle + 1 f(x)dx
= x2 1 0 0
0 1
2
= 2 – 1 = 1 unit2 = (4)(4) + ∫10(4 – 3x – x2)dx
3 3 3
 4 3 x3 1
=8+ 4x – 2 x2 – 3 0

= 8 + 34 – 3 – 1 4
2 3

= 1 0 16 unit2

13. Tentukan isi padu janaan dalam sebutan π apabila luas berlorek itu diputarkan melalui 360° pada paksi-x.

Determine the generated volume in terms of π when the shaded region is rotated through 360° about the x-axis. TP 3

CONTOH

Diberi persamaan y = 4 − x2. y = π132 – 64 + 32 2 – 1–32 + 64 – 32 24
y = 4 – x2 3 5 3 5
Given the equation y = 4 − x2.

Apabila/When = 3 4125 π unit3

x = 0 y = 4 x Kaedah Alternatif
2
Isipadu yang dijanakan

The generated volume

∫= π 2–2y2dx –2 O Oleh sebab graf ialah bersimetri pada x = 0.

= ∫π 2–2(4 – x2)2dx Isipadu dijanakan ialah
Since the graph is symmetry at x = 0, the generated
volume is

= π∫2–2(16 – 8x2 + x4)dx Tip ∫ ∫= 2 20πy2dx = 2π 20(4 – x2)2dx

 4= π 16x – 8 1 2 Guna/Use  4= 2π 16x –8 x3 + 1 x5 2
3 5 –2 3 5 0
∫baπy2dx
x3 + x5 = 34 2 π unit3
15

70

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran 

(a) y (b) y (c) y
y = x(x – 3) x = 2y2 – 4
y = 1x
Ox BAB 3

–1 O x

O1 3 x

Diberi persamaan y = 1 . Diberi persamaan y = x(x − 3). Diberi persamaan x = 2y2 – 4.
x
Given the equation y = x(x − 3). Given the equation x = 2y2 – 4.
1
Given the equation y = x . Titik persilangan dengan

Isi padu janaan paksi-x (3, 0) ,(0, 0) Apabila y = 0, x = –4
The intersection point at the
The generated volume x-axis of (3, 0), (0, 0) Isipadu dijanakan

= π∫ 3 y2dx Isipadu dijanakan The generated volume
1
∫= π 0 y2dx
3 1 The generated volume –4
= π∫ 1 x2 dx
= π∫ 3 y 2dx = 1∫π0 x + 4 2 dx
1 –4 2
3
∫= π 1 x –2dx = π∫ 3 x2(x – 3)2dx  = π1 x2 + 2x 0
1 4 –4
π– 1x 3
= 1 = ∫π 3 x 4 – 6x3 + 9x2)dx
1
= π0 – (4 – 8)4
= π– 13 –(–1)4  = π 1 x5 – 3 x4 + 3x3 3
5 2 1 = 4π unit3

= 2 π unit3 = π ( 35 – 3 (3)4 + 3(3)3) –
3 5 2
1 3
1 5 – 2 + 324

= 6 52 π unit3

14. Tentukan nilai k diberikan isipadu dalam sebutan π yang dijanakan apabila luas berlorek itu diputarkan melalui
360° pada paksi-y.

Determine the value of k with the given volume in terms of π that is generated when the shaded region is rotated through

360° about the y-axis. TP 4

CONTOH

y Isipadu dijanakan
y = 4x2
The generated volume
k

= π∫ k x2dy = 65 π
4 8

4 k 1 ydy = 65
4 8
∫ 4

Ox  4 y2 k = 65
2 2
65 4
8
Diberi isi padu janaan ialah π unit3 oleh k2 – 8 = 65
2 2
lengkung y = 4x2 dari y = 4 ke y = k.
k2 81
Given the generated volume is 65 π unit3 by the 2 = 2 k=9
8
curve y = 4x2 from y = 4 to y = k.

71

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran

(a) y (b) y (c)
4 y
y=x–3
BAB 3 k y = x2 + k k x
Ox x 1
O

O

Diberi isi padu yang dijanakan Diberi isi padu yang Diberi isi padu yang

oleh rantau berlorek ialah dijanakan oleh rantau dijanakan oleh rantau

10 π unit3 dengan graf berlorek ialah 2π unit3 berlorek ialah 152 π unit3
3 dengan lengkung y = x2 + k. 3
Given the volume generated by dengan graf y = x − 3.
y2 + 4x2 = 16 Given the volume generated by
Given the volume generated by the shaded region is 2π unit3
with the curve y = x2 + k. 152
the shaded region is 10 π unit3 the shaded region is 3 π unit3
3
with the graph y2 + 4x2 = 16. with the graph y = x − 3.

Apabila/When x = 0, y = 4 Apabila/When x = 0, y = k Isipadu = 1532π

Isipadu/Volume = 10 π Isipadu/Volume = 2π Volume
3
2π = π∫ 4 (y – k)dy = π∫1k (y + 3)2dy
4 k
= π∫ k x2dy
∫4 (y – k)dy = 2
16 – y2 k (y + 3)3 1352π
4 3
= 1π∫ 4 2 dy  y2 4 3 4π k =
k k 1
2 – ky = 2
 = 1 4= 10
π 4y – 12 y3 3 π k2 (k + 3)3 – 43 = 152
k 2 3 3 3
(8 – 4k) – ( – k2) = 2
116 64 2 k3 10
– 12 – 4k + 12 = 3 k2 (k + 3)3 = 216
2 k + 3 = 6
k3 22 – 4k + 6 = 0 k = 3
12 3
– 4k + = 0

k3 – 48k + 88 = 0 k2 – 8k + 12 = 0
(k – 2)(k – 6) = 0
Jika k = 2, 8 – 48(2) + 88 = 0 ∴k = 2 k = 6

If

∴k = 2

72

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran 

15. Tentukan isi padu yang dijanakan bagi setiap yang berikut.

Determine the volume generated for each of the following. TP 5

CONTOH (a) y BAB 3
y = 6 + 7x – 3x2
y
(2, 8)

y = (x – 2)2 y=x+4

Ox Ox

Luas yang dibatasi oleh garis lurus y = x + 4 dan Luas yang dibatasi oleh lengkung, y = 6 + 7x − 3x2
y = (x − 2)2 diputarkan melalui 360° pada paksi-x. dan pakis-y diputarkan melalui 360° pada
paksi-x.
The area bounded by the straight line y = x + 4 and
y = (x − 2)2 is rotated through 360° about the x-axis. The area bounded by the curve y = 6 + 7x − 3x2 and
y-axis is rotated through 360° about the x-axis.

Langkah 1/Step 1 Isipadu dijanakan oleh lengkung
Cari titik persilangan/Find the intersection point
(x – 2)2 = x + 4 Volume generated by the curve
x2 – 4x + 4 = x + 4
x2 – 5x = 0 = π∫20(6 + 7x – 3x2)2 dx
x(x – 5) = 0
x = 0, x = 5 = π∫20(36 + 84x + 13x2 – 42x3 + 9x4)dx

Langkah 2/Step 2 y = 164.27π atau 1 64 145π
Apabila/When x = 5, y = 9
(5, 9)
Langkah 3/Step 3 x Isipadu kon

Cari isi padu janaan lengkung. Volume of cone

Find the generated volume of the = 1 π(8)2(2)
3
curve = 42.67 π atau 42 32 π

π∫ 5 y2dx = π∫50(x – 2)4dx O
0
Isipadu janaan/Generated volume
(x – 2)5
5
  = π 5 = 121.6 π unit3
0

= 243 π + 32 π
5 5

= 275 π
5

Langkah 4/Step 4

Cari isi padu janaan garis lurus dari x = 0 ke x = 5.
Find the generated volume of the straight line from
x = 0 to x = 5.

π∫50(x + 4)2dx

(x + 4)3
3
 =π 5
0

= π1 93 – 43 2 = 665 π
3 3 3

Isi padu janaan/Generated volume

= 665 π − 275 π = 166 2 π unit3
3 5 3

73

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran

(b) y (c)

y

BAB 3 y = 4x y = 9 – x2
3 y=x
x x
A 2

O

Luas yang dibatasi oleh paksi-y, lengkung
4
y= x dan y = x diputarkan melalui 360° pada Luas yang dibatasi oleh lengkung y = 9 − x2,
tangen pada x = 2 dan paksi-x yang diputarkan
paksi-y. melalui 360° pada paksi-x.

The area bounded by the y-axis, the curve y = 4 The area bounded by the curve y = 9 − x2, the
x tangent at x = 2 and the x-axis is rotated through
and y = x is rotated through 360° about the y-axis. 360° about the x-axis.

Koordinat A/Coordinates of A y = 9 – x2

4 =x dy = –2x
x dx

x = 2, y = 2 x = 2, dy = –4
dx
Isipadu/Volume
persamaan tangen pada (2, 5)
= π∫32 x2dy + kon
The equation of tangent at (2, 5)

= π∫32 16 dy + 1 π(2)2(2) y – 5 = –4(x – 2)
y2 3
y = –4x + 13 1 13 02.
= π –16 3 + 8 π Tangen menyilang paksi-x pada 4 ,
y 2 3
1 2Tangent intersects the x-axis at 13 , 0
= π–136 –12624 + 8 4
– 3 π
1 2Isipadu kon = 1 5
= 5 13 π unit3 3 π(5)2 4
Volume of cone
= 11225π unit3

Isipadu dijanakan oleh lengkung

Volume generated by the curve

= π∫ 3 y2dx
2

= π∫32(9 – x2)2dx

= π∫32(81 – 18x2 + x4)dx

3 4= π 1 3
81x – 6x3 + 5 x5 2

= 9 15 π unit3 73
60
Isipadu diperlu = π unit3

Volume needed

74

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran 

3.4 Aplikasi Pengamiran

Application of Integration

16. Selesaikan masalah berikut yang melibatkan pengamiran. BAB 3

Solve the following problems involving integration.

CONTOH x2 = 3
3
B
AC x = ±3

Maka lebar pintu

Hence, the width of the door

7m 7m = (3 + 3)m = 6 m

(ii) Luas pintu/Area of the door

DE = ∫3 10 – x2 dx
–3 3
 2 3
3 4 = 1 –3
Rajah menunjukkan permukaan sebuah 10x – 9 x3

pintu. ABC ialah suatu lengkok simetri = 54 m2

daripada sebahagian graf y = 10 − x2 . B ialah (iii) Luas tambahan/Additional area
3
titik tertinggi dari DE. = 1 × 6 = 6 m2

The diagram shows the front view of a door. ABC Luas pintu baharu/Area of new door

is a symmetrical arch from a part of the graph = (54 + 6) m2

y = 10 − x2 . B is the highest point from DE. = 60 m2
3

(i) Cari lebar DE. 6
Find the width of DE. 1

(ii) Cari luas permukaan pintu. 8m

Find the surface area of the door.

(iii) Cari luas permukaan pintu baharu apabila

AD dan CE ditambah 1 meter, manakala

bentuk lengkok ABC dan lebar pintu

dikekalkan. Tip

Find the new surface area of the door when AD 1
and CE are lengthened by 1 metre, while the 3
y = – (x2) + 10 merupakan paksi-y yang melalui
arch ABC and the width of the door remain the
tengah-tengah pintu. Jika x = 0, y = 10 ialah titik
same.
maksimum
Penyelesaian 1
y = – 3 (x2) + 10 is a y-axis that passes the middle of
(i) Diberi/Given
the door. If x = 0, y = 10 is a maximum point.
x2
y = 10 – 3 y

Apabila/When

y=7 x2 10 m
3 O
7 = 10 –

x

75

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran

(a) Satu titik P(x, y) bergerak supaya jaraknya dari titik asalan sentiasa r unit.
A point P (x, y) moves such that it is always r units from the origin.

(i) Tunjukkan bahawa persamaan bulatan yang dibina ialah x2 + y2 = r2.
Show that the equation of the circle formed is x2 + y2 = r2.
BAB 3 4 πr3.
(ii) Seterusnya, tunjukkan bahawa isi padu sfera berjejari r unit diberi oleh 3

Then, show that the volume of the sphere with radius of r unit is given by 4 πr 3.
3

(i) Dengan teorem Pythagoras, (ii) Isi padu janaan/Generated volume

x2 + y2 = r2 y = 2π∫ r x2dy = 2π 3r3 – r3 4
0 3

= 2π∫ r –
P(x, y)  4 0 (r2 y2)dy = 2π 3 2 r34 = 4 πr 3
3 3
ry y3 r
3 0
= 2π r2y –

Ox x

(b) Dengan cara yang sesuai, buktikan bahawa isi padu kon ialah 1 πr2h, dengan keadaan r ialah jejarinya
3
dan h ialah tinggi kon itu.
1
With a suitable method, prove that the volume of a cone is 3 πr 2h, where r is the base radius and h is the height of
the cone.

Persamaan garis lurus π∫h r(y – h) 42dy = r2π 30 – 1 –h3 24
= 0 h h2 3
The equation of straight line

y= h x + h = r2π ∫h0(y – h)2dy 3 4= r2π h3 = 1 πr2h unit3
r h2 h2 3 3
Isipadu dijana/Generated volume
r2π (y – h)3 h
h2 30
= π∫ h0x2dy  4=

(c) Satu kon aiskrim wafer dengan tinggi 10 cm dan jejari tapak ialah 3 cm

akan dipenuhi oleh aiskrim sehingga bahagian atasnya berbentuk parabola Aiskrim 3 cm
4 Ice cream
9
dengan persamaan y = 14 − x2 seperti ditunjukkan dalam rajah. Cari

A wafer ice cream cone is 10 cm tall and a base radius of 3 cm will be filled with ice
4
cream until the top is part of a parabola with the equation y = 14 − 9 x2 as shown in 10 cm
the diagram. Find

(i) isi padu aiskrim bagi setiap kon.

the volume of ice cream for each cone.

(ii) keuntungan seorang penjual jika setiap kon boleh dijual dengan harga RM3.50 dan dia boleh menjual

dua bekas aiskrim dengan dimensi (30 × 20 × 40) cm. Kos setiap bekas aiskrim ialah RM345.

the profit of the seller if each cone is sold at RM3.50 and he can sell two containers of the ice cream whose
dimensions are (30 × 20 × 40) cm. The cost of each container of the ice cream is RM 345.

y = 14 – 4 x2 Isipadu kon/Volume of cone = 1 π(3)2(10) = 30π
9 3
14
∫π 10 x 2dy Jumlah isipadu/Total volume = 18π + 30π = 48π unit3

= 9 ∫π 1140(14 – y)dy Satu bekas isipadu = 20 × 30 × 40 = 24000 cm3
4
Volume of a container

3 4=9 π 14y – y2 14 Dua bekas/Two containers = 48000 cm3
4 2 10
Bilangan kon = 48000 = 318 kon
9 48π
= 4 π [(196 – 98) – (140 – 50)] Number of cone

9 Jumlah jualan/Total sales = 318 × RM3.50 = RM1113
4
= π [8] = 18π unit3 Kos/cost = RM345 × 2 = RM690

Keuntungan/Profit = RM423

76

PRAKTIS   Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran 

SPM 13

Kertas 1 3. Rajah menunjukkan graf suatu lengkung y = f(x) BAB 3
dan garis lurus y = g(x).
1. Diberi ∫ 4 f(x)dx = 5, cari
–2 The diagram shows a graph for the curve y = f(x) and
the line y = g(x).
Given that 4 f(x)dx, find
∫2014 –2 y

(a) ∫ –42f(x)dx

(b) ∫ 4–2[2f(x) – 3x]dx y = g(x)
y = f(x)
(a) ∫ 4 f(x)dx = 5 x
–2
h
∫ –2 f(x)dx = –5 (0, 3k)
4 –h O

 4(b) ∫ 4 2f(x)dx –∫ 4 3x dx
–2 –2
x2
3 4
= 10 – 2
–2
Diberi h f(x)dx – h g(x)dx = 8.
= 10 – [24 – 6] = –8
∫0 h ∫0 h
0 0
f(x)dx – g(x)dx = 8.
∫ ∫Given that
(a) Pada rajah, lorekkan rantau yang diwakili
h h
oleh f(x)dx – g(x)dx.
∫0 ∫0
On the diagram, shade the region represented by

∫ ∫h h
0 0
f(x)dx – g(x)dx.

(b) Cari h f(x)dx dalam sebutan h dan k.

∫0
h
2. Fungsi kecerunan suatu lengkung ialah (2x – 4). ∫Find 0 f(x)dx in terms of h and k.

Diberi P(3, –2) terletak pada lengkung itu, cari (a) y

2015 The gradient function of a curve is (2x – 4). Given that (0, 3k) (h, 6k)
P(3, –2) lies on the curve, find
–h O h x
(a) kecerunan tangen pada titik P.

the gradient of the tangent at P.

(b) persamaan lengkung itu.

the equation of the curve.

(a) dy = 2x –4
dx

Apabila x = 3, dy = 2(3) – 4 (b) ∫h f(x)dx = 8 + ∫h g(x)dx
dx 0 0

=2 =8+ 1 [3k + 6k]h
2
(b) y = ∫(2x – 4)dx
9hk
= x2 – 4x + c = 8 + 2

Pada/At (3, –2), –2 = 9 – 12 + c

c = 1

∴y = x2 – 4x + 1

77

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran 6. y

4. Diberi ∫h–1(3 – 2x)dx = 6, cari nilai yang mungkin 2018 O
y = –4
BAB 3 bagi h. x

∫2016 Given that h–1(3 – 2x)dx = 6, find the possible values of h. Rajah di atas menunjukkan lengkung y = f(x).
Garis lurus y = –4 ialah tangen kepada lengkung.
∫h–1(3 – 2x)dx = 6 Diberi f'(x) = 4x – 4, cari persamaan lengkung itu.

[3x – x2]h–1 = 6 Diagram above shows a curve y = f(x). The straight line
3h – h2 – (–3 – 1) = 6 y = –4 is a tangent to the curve. Given that f'(x) = 4x – 4,
3h – h2 – 2 = 0 find the equation of the curve.
h2 – 3h + 2 = 0
(h – 1)(h – 2) = 0 f'(x) = 4x – 4
h = 1; 2 f'(x) = 4x – 4 = 0
x=1
∴(1, –4) ialah titik minimum
5. Rajah menunjukkan sebahagian lengkung y = f(x).
∫b (1, –4) is a minimum point
Rantau berlorek ditakrifkan sebagai a f(x)dx = 8.
f(x) = ∫(4x – 4)dx
2017 The diagram shows a part of the curve y = f(x). The
b = 2x2 – 4x + c
∫shaded region is defined by a f(x)dx = 8. Pada/At (1, –4)
–4 = 2 – 4 + c
y c = –2
f(x) = 2x2 – 4x – 2
y = f(x)

–2 O 1 3 5 x

(a) Nyatakan nilai a dan nilai b. 3
State the value of a and of b. 4
7. Diberi ∫ h p(x)dx = , cari
(b) Diberi bahawa luas dibatasi oleh lengkung, 2

paksi-x dan x = –2 ke x = 3 ialah 13 unit2. Cari ∫2019 h p(x)dx 3
3 Given that 2 = 4 , find
–2
nilai ∫ f(x)dx. (a) ∫ 2 p(x)dx
h
Given that the area bounded by the curve, the x-axis
and x = –2 to x = 3 is 13 unit2. Find the value of 1
(b) ∫ h [p(x) – 2x]dx = –4 4
∫3 f(x)dx. 2
–2
3
(a) ∫ b f(x)dx = 8 (a) ∫ 2 p(x)dx = – 4
a h

a = 1, b = 3 (b) ∫ h p(x)dx – ∫ h 2xdx = –4 1
2 2 4
( b) ∫ 3 f(x)dx = 1 f(x)dx + 3 f(x)dx
–2 3 +4 1 = [x2]h2 = h2 – 4
∫ –2 ∫1 4 4

= –5 + 8 = 3

5 + 4 = h2

h = 3

78

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran 

Kertas 2 1 2(c) π∫0k 110
3
1. Rajah menunjukkan lengkung y = 3x2 – 27 dan AB
ialah tangen kepada lengkung itu pada titik P. Isipadu = x2 dy = 2 π

2017 The diagram shows the curve y = 3x2 – 27 and AB is a Volume
tangent to the curve at point P.
∫ 01 1 y + 92dx = 220 BAB 3
y 3 3
k

 4 1 y2 + 9y 0 = 220
6 3
k

0 – [ k2 + 9k] = 220
6 3

A k2 + 54k + 440 = 0

Ox (k + 44)(k + 10) = 0

P k = –44 ; –10

B ∴k = –10

Diberi bahawa kecerunan tangen ialah 6. 2. Rajah menunjukkan garis lurus 4y = x + 3
menyentuh lengkung x = 4y2 – 2 pada titik A.
Given that the gradient of the tangent is 6.
2018 The diagram shows a straight line 4y = x + 3 touches the
(a) Cari koordinat P. curve x = 4y2 – 2 at point A.

Find the coordinate of P. y

(b) Hitung luas rantau berlorek itu. A x = 4y2 – 2
Ox
Calculate the area of the shaded region.

(c) Apabila rantau yang dibatasi oleh lengkung,
paksi-x dan y = k diputarkan melalui
180° pada paksi-y menghasilkan isi padu

3 6 32 π unit3, cari nilai k.

When the area bounded by the curve, the x-axis
and y = k is rotated through 180° about the y-axis
produces a volume of 36 23 π unit3, find the value
of k.

(a) y = 3x2 – 27 Cari / Find
(a) koordinat A.
Apabila y = 0
the coordinates of A.
x = ±3
(b) luas rantau berlorek.
Addyxpa=bi6lax ddyx = 6x = 6
the area of the shaded region.
x = 1
(c) isi padu kisaran dalam sebutan π apabila
y = –24 rantau yang dibatasi oleh lengkung dan x = 1
diputarkan melalui 180° pada paksi-x.
∴P[1, –24]
the generated volume in terms of π when the area
(b) Persamaan tangen y + 24 = 6(x – 1) bounded by the curve and x = 1 is rotated through
180° about the x-axis.

Equation of tangent (a) 4y = x + 3 ; x = 4y2 – 2

y = 6x – 30 4y – 3 = 4y2 – 2

luas berlorek 4y2 – 4y + 1 = 0

shaded area
1 (2y – 1)2 = 0
= 2 [4][24] – 3 ydx
1
∫1 y= 2 , x = 2 – 3 = –1

= 48 – ∫31(3x2 – 27)dx A–1, 1 2
2
= 48 – [x3 – 27]31
∴48 – [0 + 26] = 22 unit2

79

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran

(b) Apabila/When y = 0, x = –3 y (a) Tunjukkan bahawa a = 4.
luas berlorek/shaded area
Show that a = 4.
 2 =1 (2) 1 ∫– –1 y dx ΂–1, 21΃
2 2 –2 (b) Hitung luas berlorek itu.

Calculate the shaded area.

BAB 3 1 1 1
2 2
= ∫– –1 (x + 2) 2 dx x (a) y = 6x – x2
–2
–3 –2 –1 O ddxy = 6 – 2x = 0
x = 3
3 4 =1 – 1 . 2 (x + 2) 3 –1
2 2 3 2 –2

= 1 – 3 1 – 04 = 1 unit2 Katakan persamaan tangen
2 3 6
Let the equation of tangent

1 y = mx + c
–2
(c) Isipadu/Volume = ∫π y2dx 10 = 3m + c ...➀

1 1 6x – x2 = mx + c
–2 4
= ∫π (x + 2)dx x2 + (m – 6)x + c = 0

(x + 2)2 (m – 6)2 – 4c = 0
8
3 4 = π 1 (m – 6)2 = 4(10 – 3m)
–2
m2 – 12m + 36 = 40 – 12m
π3 9 – 04 = 9π
= 8 8 m2 = 4

Isipadu sebenar m = ±2

Actual volume Apabila m = –2, c = 10 – 3(–2)

= 9π unit3 = 16
16
∴ 6 – 2x = –2

8 = 2x

x = 4; y = 6(4) – 16

= 8

∴A(4, 8) ; a = 4

(b) luas berlorek = luas trapezium – 4 ydx

∫3

shaded area = area of trapezium – 4 ydx
∫ 3

1 + 8][1] ∫4 (6x – x2)dx
3 4 = 2 [10 – 3

– 1 4
= 3 3
9 – 3x2 x3

= 9 – 326 2 – 184 = 1 unit2
3 3

3. Rajah menunjukkan lengkung y = 6x – x2 dan y
tangen kepada lengkung pada titik A melalui titik (3, 10)
(4, 8)
2019 B(3, 10).
O 34 x
The diagram shows a curve y = 6x – x2 and the tangent
to the curve at point A passes through the point B(3, 10).

y
B(3, 10)
A

O 3a x Praktis
SPM
Ekstra

80

  Matematik Tambahan  Tingkatan 5  Bab 3 Pengamiran 

Sudut KBAT KBAT

Ekstra

Rajah menunjukkan sebahagian graf y = x3 + 1. Tentukan / Determine BAB 3
P ialah titik persilangan lengkung pada paksi-x. Satu (a) koordinat P dan Q.
tangen kepada lengkung di P memotong lengkung
sekali lagi di Q. QR adalah berserenjang dengan the coordinates of P and Q.
paksi-x.
(b) nisbah luas A kepada luas B.
The diagram shows part of a graph y = x3 + 1. P is the
intersection point of the curve at the x-axis. A tangent to the the ratio of the area of A to B.
curve at P cuts the curve again at Q. QR is perpendicular to
the x-axis. (c) isi padu janaan apabila luas berlorek A
diputarkan melalui 180° pada paksi-x.
y
the generated volume when the shaded area A is
Q rotated through 180° about the x-axis.

P B x
O A
R

y = x3 + 1

(a) y = x3 + 1 = (4 + 2) – 1 1 – 12
4
Apabila y = 0 ; 0 = x3 + 1

x3 = –1 =6+ 3
4
x = –1

P[–1, 0] = 6 34 unit2

dy = 3x2 1 27
dx 2 2
dy luas/area PQR = (3)9 = unit2
Apabila x = –1, dx = 3
When 27 6 34 = 6 43
∴luas/area B = 2 – unit2
y = 3(x + 1)
B = 6 34 : 6 43
Jika y = 3x + 3 menyilang lengkap Nisbah A :
Ratio
If y = 3x + 3 intersects completely

3x + 3 = x3 + 1 =1:1

x3 – 3x – 2 = 0 π 2 π ∫2–1(x3 1)2dx
3 4(c) Isipadu =2 y2dx = 2 +
Apabila x = 2 23 – 3(2) – 2 ∫ –1
Volume
=8–8=0 = π 1 x7 + 1 x4 + x 2
2 7 2 –1
∴y = 23 + 1 = 9

Q(2, 9) y = π 32 8 1143 4
Q 2

(b) luas A = ∫2 ydx (2, 9) = 14 2183 π unit3
area A = –1

∫2–1(x3 + 1)dx

3 4 = x4 +x2 P Rx
4 –1 –1O 2

Quiz 3

81

BAB 4 BAB  Matematik Tambahan  Tingkatan 5  Bab 4 Pilih Atur dan Gabungan

4 Pilih Atur dan Gabungan
Permutation and Combination

4.1 Pilih Atur

Permutation

NOTA IMBASAN
NOTA IMBASAN

1. Petua pendaraban: 5. Pilih atur bagi n objek yang berlainan dengan syarat
tertentu.
Multiplication rule:
Permutations of n different objects according to some conditions.
Jika suatu peristiwa boleh berlaku dalam m cara dan Bagi masalah yang melibatkan pilih atur objek dengan

suatu peristiwa kedua boleh berlaku dalam n cara, maka syarat tertentu, pastikan syarat tertentu dipenuhi
terlebih dahulu sebelum menyusun yang lain.
kedua-dua peristiwa boleh berlaku dalam m × n cara. For problems involving permutations of the objects for given
conditions, make sure that the certain given conditions need to
If an event can happen in m ways and another event can happen be addressed first before permuting the rest.
6. Jika terdapat n benda, antaranya ada p benda yang
in n ways, then the two events can happen successively in m × n serupa, q benda yang lain yang serupa juga, r benda

ways.

2. Jika n benda yang berlainan disusun dalam satu baris

mengikut tertib, bilangan pilih atur

If n different objects are arranged in a line according to order, the

number of permutations lain lagi yang serupa lagi dan seterusnya, maka jumlah

= n! = n(n − 1)(n − 2) …. 3 . 2 . 1 pilihatur ialah n!
p!q!r!...
Tatatanda n! dibaca ‘n faktorial’.

The notation n! is read as ‘factorial n’. If there are n objects, among which are p objects that are similar,

3. Bilangan pilih atur untuk menyusun 0 objek ditandakan another group of q objects that are similar to each other, another

0! = 1. group of r objects that are similar to each other, then the total

The number of permutations to arrange 0 object is labelled as permutation is n!
p!q!r!...
0! = 1.

4. Jika r objek dipilih daripada n objek yang berlainan 7. Jika terdapat n objek yang berlainan disusun dalam

dan disusun mengikut tertib, bilangan pilih atar ialah satu bulatan, maka bilangan pilih atur ialah (n − 1)! jika

nPr = (n n! . posisi tidak ditetapkan pada bulatan itu.
– r)! If there are n different objects to be arranged in a circle, then the

If r object is chosen from n different objects and are arranged in permutation is given by (n − 1)! if the positions are not marked in
the circle.
order, the number of permutations is nPr = n! .
(n – r)!

1. Nilaikan yang berikut.

Evaluate the following. TP 1

CONTOH

(i) nPn (ii) 6P2 (iii) 7P0 (ii) 6P2 = (6 6!
– 2)!
Penyelesaian:
n 6!
(i) nPn = (n – n)! = 4!

n! Guna /Use = 6.5.4.3.2.1 Tip
(0)! 4.3.2.1
= nPr = n! (n – r)! ≠ n! – r!
(n – r)!
= 30
= n!

82

  Matematik Tambahan  Tingkatan 5  Bab 4 Pilih Atur dan Gabungan 

7!
– 0)!
(iii) 7P0 = (7 Sudut Kalkulator

= 7! (i) 7P4
7! Tekan
Press 7 SHIFT nPr 4 = 840 BAB 4
7.6.5.4.3.2.1
= 7.6.5.4.3.2.1 (ii) 5!

= 1 Tekan 5 SHIFT X! = 120
Press

(a) (i) nP0 (b) (i) 10P3 (c) (i) nPn − 1 × n − 1P2
(ii) nPn − 1 (ii) 5P4 × 6P2 (ii) mm+P1P2 2

(i) (n n! = 1 (i) 10! = 720 (i) n! × (n – 1)!
– 0)! (10 – 3)! (n – n + 1)! (n – 1 – 2)!

(ii) (n – n! 1)! = n! ( ii) (51)!! × 6! = 5 × 6! = n!(n – 1)(n – 2)
n+ (4)! = 3 6000
m! (m + 1)!
(ii) (m – 2)! ÷ (m + 1 – 2)!

= m(m – 1) ÷ (m + 1)!
(m – 1)!

= m–1
m+1

2. Selesaikan masalah yang berikut. (a) Terdapat 4 jenis sayur dan 3 jenis buah-buahan
yang dijual di sebuah gerai yang berdekatan
Solve the following problems. TP 2 dengan rumah John. Cari bilangan cara John
boleh membeli 1 jenis sayur dan 1 jenis buah-
CONTOH buahan.

Gopal ingin memilih sebuah buku dan sebuah There are 4 types of vegetables and 3 types of fruits
majalah dari rak yang mengandungi 3 buah buku being sold at a store near to John’s house. Find the
dan 2 buah majalah yang berlainan. Berapa cara number of ways of John to buy 1 type of vegetable
berlainan yang boleh dilakukan oleh Gopal? and 1 type of fruit.

Gopal wants to choose a book and a magazine from Bilangan cara
a shelf which has 3 different books and 2 different
magazines. How many ways can this be done by Gopal. Number of ways

Penyelesaian: =4×3
Gopal mempunyai 3 cara untuk memilih sebuah = 12
buku dan 2 cara untuk memilih sebuah majalah
dari rak itu.
Mengikut petua pendaraban, jumlah cara ialah
3 × 2 = 6.

Gopal has 3 ways to choose a book and 2 ways to
choose a magazine from the shelf.
By using multiplication rule, the total number of ways
is 3 × 2 = 6.

83

BAB 4   Matematik Tambahan  Tingkatan 5  Bab 4 Pilih Atur dan Gabungan (c) Dalam satu pameran perabot, ada 6 jenis
kerusi plastik, 4 jenis meja kecil dan 3 jenis
(b) Seorang murid ternampak 5 batang krayon kusyen. Puan Wan ingin memilih satu barang
dan 6 batang pensel berwarna di atas lantai. daripada setiap jenis barang itu. Cari bilangan
Guru kelas menyuruh murid itu memilih cara berlainan yang boleh dilakukan oleh
sebatang krayon dan pensel berwarna, cari Puan Wan?
berapa cara yang boleh dilakukan oleh murid
itu In a furniture exhibition, there are 6 types of plastic
chairs and 4 types of small tables and 3 types of
A student saw 5 sticks of crayons and 6 colour cushions. Miss Wan wants to select one from each
pencils on the floor. The class teacher wants the group of items. Find the number of ways this can be
student to pick up a crayon and a colour pencil, find done by Miss Wan.
the number of ways that can be done by the student.
Bilangan cara
Bilangan cara
Number of ways
Number of ways
=6×4×3
=5×6 = 72
= 30

3. Selesaikan masalah pilih atur n objek yang berbeza. (a) Seorang pekerja diminta menyusun 7 bahan
berlainan penghasilan kilang dalam satu baris
Solve the permutation problems of n different objects. TP 2 untuk disemak. Berapa cara boleh dilakukan?

CONTOH A worker is asked to arrange 7 different materials
produced by a factory in a line to be checked. In how
Berapa cara berlainan digit 0, 1, 2, 3 dan 4 boleh many ways can this be done?
disusun dalam satu baris?
Bilangan cara = 7! =5 040
How many different ways of the digits 0, 1, 2, 3 and 4
can be arranged in a line? Number of ways

Penyelesaian:

Terdapat lima tempat kosong perlu diisi oleh

nombor 0, 1, 2, 3 atau 4.

Tempat pertama ada 5 cara, tempat kedua ada

4 cara kerana satu digit telah digunakan, tempat

ketiga ada 3 cara kerana dua digit telah digunakan,

tempat keempat ada 2 cara kerana tiga digit telah (b) Terdapat 8 keping gambar perlu disimpan
dalam sebuah buku album. Berapa cara ini
digunakan. Tempat terakhir hanya ada 1 cara boleh dilakukan?

kerana satu digit yang tertinggal sahaja. There are 8 photos to be arranged in a photo album.
In how many ways can this be done?
Maka dengan petua pendaraban, bilangan cara =
Bilangan cara = 8! = 40 320
5×4 × 3 × 2 × 1 = 120 atau 5! failtleadub5yP5th. e numbers
Number of ways
There are five ways that need to be
0, 1, 2, 3 or 4.
The first place has 5 ways, the second place has 4 ways

because one digit has been used, the third place has
3 ways because two digits have been used, the fourth
place has 2 ways because three digits have been used.
The last place has only 1 way because only one digit

is left.

So, with multiplication tips, the number of ways =
5 × 4 × 3 × 2 × 1 = 120 or 5! or 5P5.

84

  Matematik Tambahan  Tingkatan 5  Bab 4 Pilih Atur dan Gabungan  BAB 4

(c) 6 buah beg akan digantung pada 6 penyangkut dalam satu baris. Cari bilangan cara ini boleh dilakukan.

6 bags are to be hung on 6 hangers in a row. Find the number of ways this can be done.

Bilangan cara = 6! = 720

Number of ways

4. Selesaikan masalah pilih atur n objek yang berbeza diambil r objek pada satu masa.

Solve the problems of permutations of n different objects taken r objects at a time. TP 2

CONTOH (a) Terdapat 4 buah kerusi sahaja di ruang

Satu kata laluan terdiri daripada 6 digit dengan menunggu tetapi ada 10 orang mencari tempat
menggunakan digit 0, 1, 2, 3, 4, 5, 6, 7, 8 dan 9
tanpa ulangan. Cari bilangan cara berlainan kata duduk. Berapakah bilangan cara tempat
laluan ini boleh dilakukan.
duduk ini dapat diduduki?
A password is made up of 6 digits taken from the digits
0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 without repetition. Find the There are 4 chairs in a waiting area but there are
number of different passwords that can be formed. 10 people looking for the seats. How many ways can
the people be seated?
Penyelesaian:
Bilangan cara = 10P4 = 10 × 9 × 8 × 7
Number of ways = 5 040

Terdapat lima tempat kosong perlu diisi oleh

10 nombor.

There are five empty spaces to be filled by 10 numbers.

Tempat pertama ada 10 pilihan.

The first space has 10 choices.

Tempat kedua ada 9 pilihan kerana satu digit

telah digunakan. (b) Seorang pelukis mempamerkan 9 lukisan di
papan pamerannya. Encik Tan ingin membeli
The second space has 9 choices since one digit has been tiga lukisan untuk rumah baharunya. Cari
used. bilangan cara yang boleh dilakukan.

Tempat ketiga ada 8 pilihan kerana dua digit An artist exhibits 9 drawings on an exhibition board.
Mr.Tan wants to buy three paintings for his new
telah digunakan. house. Find the number of ways that can be done.

The third space has 8 choices since two digits have Bilangan cara = 9P3
Number of ways = 504
been used.

Tempat keempat ada 7 pilihan kerana tiga digit

telah digunakan.

The fourth space has 7 choices since three digits have

been used.

Tempat kelima ada 6 pilihan kerana empat digit

yang tertinggal sahaja.

The fifth space has 6 choices since four digits have been
used.

Tempat terakhir masih ada 5 pilihan.

The final space has 5 choices.

Maka, dengan petua pendaraban, bilangan cara

So, by multiplication rule, the number of ways

= 10 × 9 × 8 × 7 × 6 × 5 = 151 200

atau/or 10P6 = 10! = 151 200
(10−6)!

85

BAB 4   Matematik Tambahan  Tingkatan 5  Bab 4 Pilih Atur dan Gabungan

(c) Daripada 6 pakej yang ditawarkan oleh Astro, bapa Hassan ingin membeli 3 pakej itu. Cari bilangan
cara yang boleh dilakukan.

From the 6 packages offered by Astro, Hassan’s father wants to purchase 3 packages. Find the number of ways that
can be done.

Bilangan cara = 6P3
Number of ways = 120

5. Selesaikan setiap soalan yang berikut yang melibat objek secaman.

Solve each of the following questions involving similar objects. TP 3

CONTOH (a) Terdapat 4 batang pen biru yang sama, 3 batang

Cari bilangan cara untuk menyusun semua huruf pen merah yang sama dan 2 pen hitam yang

daripada perkataan KALKULATOR. sama disusun dalam satu baris. Cari bilangan

Find the number of ways to arrange all the letters in the cara untuk menyusun semua pen itu.
word KALKULATOR.
There are 4 blue pens of the same type, 3 red pens
Penyelesaian: that are the same and 2 black pens that are also the
same to be arranged in a row. Find the number of
Jika semua huruf adalah berlainan, maka ways to arrange all the pens.

terdapat 10! cara untuk menyusun semua huruf Bilangan cara susunan

itu. Tetapi terdapat 2 huruf K yang sama, 2 huruf Number of ways of arrangement

A yang sama dan 2 huruf L yang sama. Maka = 9! = 1 260
4!3!2!
bilangan cara akan berkurang kerana tidak dapat

mengasingkan huruf yang sama.

Jadi, bilangan cara susunan

If all the letters are different, then there are 10! ways to
arrange all the letters. But there are 2 similar letters of K,
2 similar letters of A and 2 similar letters of L. Hence the

number of ways will decrease because cannot separate

the same letters. So, the number of ways of arrangement
10!
= 2!2!2! = 453 600

(b) Dalam sebuah kedai yang menjual telefon (c) Rajah di bawah menunjukkan 9 kad yang
bernombor.
bimbit, terdapat 5 model A, 6 model B dan
Diagram below shows 9 numbered cards.
4 model C. Cari bilangan cara pekerja dapat
222566688
mempamerkan ketiga-tiga model itu di dalam Cari bilangan cara menyusun semua kad ini

kedai. dalam satu baris.

In a shop that is selling smart phones, there are Find the number of ways to arrange all the cards in a
5 model A, 6 Model B and 4 model C. Find the row.
number of ways of the staff to exhibit all the three
models in a shop.

Bilangan cara susunan Bilangan cara susunan

Number of ways of arrangement Number of ways of arrangement

= 15! = 630 630 = 9! = 5 040
5!6!4! 3!3!2!

86

  Matematik Tambahan  Tingkatan 5  Bab 4 Pilih Atur dan Gabungan 

6. Selesaikan soalan berikut yang melibatkan susunan objek dalam bulatan.

Solve the following questions involving arrangement of objects in a circle. TP 3

CONTOH (a) Dalam suatu permainan, seramai 8 orang BAB 4
kanak-kanak perlu duduk di atas lantai dalam
Dengan berapa cara 7 orang boleh duduk suatu bulatan. Cari bilangan cara menyusun
mengelilingi sebuah meja bulat? semua kanak-kanak itu.

In how many ways can 7 people be seated around a In a game, 8 children are to be seated in a circle on
round table? the floor. Find the number of ways to arrange the
children.
Penyelesaian:
Oleh sebab meja bulat, bilangan cara ialah Bilangan cara = 7!
(7 − 1)! = 720 Number of ways = 5 040

Since it is a round table, the number of ways is (7 − 1)!
= 720

Tip

Jika objek di sebelah kiri dan kanan adalah sama,

susunan itu adalah sama. Oleh itu, dalam kes ini,
7!
7 mungkin telah hilang, iaitu 7 = 6!

If the objects on the left and on the right are the

same, it is the same arrangement. Hence, in this
7!
case, 7 may have been lost, that is 7 = 6!

(b) Terdapat sebuah pokok di taman. Pekebun (c) Dalam satu jamuan makan malam, terdapat
ingin menyusun 6 pasu bunga mengelilingi 10 orang akan duduk pada satu meja bulat.
pokok itu. Cari bilangan cara dia dapat Cari bilangan cara menyusun semua 10 orang
menyusun pasu bunga tersebut. tersebut.

There is a tree in a garden. A gardener wants to put In a dinner, there are 10 people to be seated around a
6 pots of flowers around the tree. Find the number of round table. Find the number of ways to arrange all
ways of the gardener to arrange the flower pots. the 10 people.

Bilangan cara = 5! Bilangan cara = (10 − 1)!
Number of ways = 120 Number of ways = 362 880

7. Selesaikan setiap yang berikut. (ii) bermula dengan 5.

Solve each of the following. TP 4 starts with 5.

CONTOH (iii) berakhir dengan nombor genap.

Berapa banyak nombor 4 digit dapat dibentuk ends with an even number.
daripada digit-digit 1, 2, 3, 4, 5, 6, 7 dan 8 jika
ulangan tidak dibenarkan dan setiap nombor (iv) bernilai lebih daripada 3 000 dan adalah

How many 4-digit numbers can be formed from nombor genap.
the digits 1, 2, 3, 4, 5, 6, 7 and 8 if repetition is not
allowed and each number greater than 3 000 and are even numbers.

(i) ialah nombor ganjil.

is an odd number.

87

  Matematik Tambahan  Tingkatan 5  Bab 4 Pilih Atur dan Gabungan

Penyelesaian: (iii)

(i) 1, 3, 5, atau 2,4,6 atau 8
7

BAB 4 Nombor genap mesti berakhir dengan digit

Nombor ganjil mesti berakhir dengan digit nombor genap.

nombor ganjil An even number must be ended with a digit of even

number.

An odd number must end with an odd digit Bilangan cara mengisi tempat terakhir

number. ialah 4. Selepas itu, masih tertinggal

Bilangan cara mengisi tempat terakhir 7 nombor yang belum digunakan untuk

ialah 4. mengisi 3 tempat kosong itu.
Number of ways to fill the last place is 4.
The number of ways to fill the last place is 4. After
Selepas itu, masih tertinggal 7 nombor that, there are left 7 numbers that have not been used
to fill the 3 empty spaces.
yang belum digunakan untuk mengisi
Cara mengisi ialah 7P3
3 tempat kosong itu.
After that, there are still 7 numbers that have The ways to fill is 7P3.

not been used to fill the 3 empty spaces. JToutmal lnauhmcbaerraoifawlaahys==44××77PP33 = 840

Cara mengisi ialah 7P3. = 840

The way to fill is 7P3. (iv) Boleh di isi

Jumlah bilangan cara = 4 × =7P834=0 840 oleh 3, 4, 5, 2, 4, 6 atau 8

Total number of ways = 4 × 7P3

6, 7 atau 8

Tip Kotak di atas menunjukkan semua nombor

Syarat perlu dipertimbangkan dahulu. yang mungkin bagi tempat pertama dan
Selepas itu gunakan petua pendaraban.
The conditons need to be attended first. terakhir.

Then, use the multiplication rule. The boxes above shows all the possible digits to be

used for the first and the last spaces.

Kes I / Case 1

Jika tempat pertama menggunakan 3, 5 atau

(ii) 7, maka tempat terakhir mempunyai 4 pilihan

5 dan terdapat 6P2 untuk dua tempat di tengah-
tengah.
If the first place using 3, 5 or 7, then the last place has

Nombor 5 mesti diletakkan pada tempat 4 choices and 6P2 for the two middle places.

pertama. Bilangan cara mengisinya ialah Bilangan cara ialah 3 × 6P2 × 4 = 360

1. Number of ways is 3 × 6P2 x 4 = 360

The number of 5 must be put at the first place. Kes 2 / Case 2
The number of ways to fill it is 1.
Jika tempat pertama menggunakan 4, 6 atau
Selepas itu, masih tertinggal 7 nombor
8, maka tempat terakhir mempunyai 3 pilihan
yang belum digunakan untuk mengisi
dan terdapat 6P2 untuk dua tempat di tengah-
3 tempat kosong itu. tengah.
If the first place uses 4, 6 or 8, then the last place has
After that, there are still left 7 numbers that
have not been used to fill the 3 empty spaces. 3 choices and there is 6P2 choices for the middle two
places.
Cara mengisi ialah 7P3.

The ways to fill is 7P3. Bilangan cara ialah 3 × 6P2 × 3 = 270
Number of ways is 3 × 6P2 × 3 = 270
Jumlah bilangan cara = 1× =7P231=0 210 Jumlah cara ialah = 360 +270 = 630

Total number of ways = 1 × 7P3

Total number of ways = 360 + 270 = 630

88

  Matematik Tambahan  Tingkatan 5  Bab 4 Pilih Atur dan Gabungan 

(a) Seorang pelajar dijemput menyusun semua (b) Ketua pengawas dan 4 orang pelajar lain yang BAB 4
huruf daripada perkataan KETUA dalam satu terdiri daripada 2 lelaki dan 2 perempuan
baris. Cari bilangan cara untuk menyusun perlu disusun dalam satu baris dalam sesi
huruf jika bergambar, Cari bilangan cara menyusun
semua 5 pelajar itu jika
A student is asked to arrange all the letters from the
word KETUA in a row. Find the number of ways to A head prefect and 4 other students consisting of
arrange the letters if 2 boys and 2 girls are to be arranged in a row in a
photo session. Find the number of ways to arrange
(i) susunan mesti bermula dengan konsonan. all the 5 students if

the arrangement must start with a consonant. (i) ketua pengawas mesti berada di tengah-
tengah baris.
(ii) konsonan mesti berada sebelah-
menyebelah antara satu sama lain. the head prefect must be in the middle of the
row.
the consonants must be side-by-side with each
other. (ii) ketua pengawas berada di tengah-tengah
dan 2 orang lelaki mesti bersama.
(iii) semua vokal mesti diasingkan.
the head prefect must be in the middle and the
all the vowels must be separated. 2 boys must be together.

(i) Bilangan cara = 2 × 4! = 48 (i) Bilangan cara = 1 × 4! = 24
Number of ways
Number of ways (ii) Bilangan cara = 1! × 2! × 2! × 2! = 8
Number of ways
(ii) Bilangan cara = 4! × 2! = 48
Number of ways
(iii) Susunan dalam bentuk VKVKV
Arrangement in the form of VKVKV
Bilangan cara = 2! × 3! = 12

Number of ways

(c) 4 orang lelaki, 2 orang perempuan dan seorang (d) Diberi digit-digit 1, 2, 3, 4, 5 dan 6 untuk
kanak-kanak diminta duduk dalam satu baris. membentuk 5 digit nombor. Jika semua digit
Cari cara menyusun mereka jika boleh diulangi, cari bilangan cara membentuk

4 boys, 2 girls and a child are asked to sit in a row. Given the digits 1, 2, 3, 4, 5 and 6 to make a 5 digits
Find the number of ways to arrange them if number. If all the digits can be repeated, find the
number of ways to form
(i) kanak-kanak duduk di antara dua
perempuan itu. (i) nombor ganjil.

the child sits between the two girls. odd numbers.

(ii) 4 orang lelaki mesti duduk bersama. (ii) suatu nombor dengan digit genap mesti
berada di tempat ganjil.
the 4 boys must sit together.
a number with the even digits must be at the
(iii) 2 perempuan tidak boleh duduk bersama. odd places.

the 2 girls cannot sit together. (i) 6 × 6 × 6 x 6 × 3 = 3 888
(ii) 3! × 3 × 2 = 36
(i) 5! × 2 = 240
(ii) 4! × 4! = 576
5! × 6P2 = 3 600

89

  Matematik Tambahan  Tingkatan 5  Bab 4 Pilih Atur dan Gabungan

8. Selesaikan setiap yang berikut.

Solve each of the following. TP 4

BAB 4 CONTOH (a) 10 orang pengakap terdiri daripada 6 orang
lelaki dan 4 orang perempuan duduk
Dengan berapa cara boleh menyusun 6 orang mengelilingi sebuah unggun api dalam satu
bulatan. Cari bilangan cara menyusun mereka
murid, dengan keadaan dua orang ialah anak jika

kembar di satu meja bulat? 10 scouts which is consisted of 6 boys and 4 girls
sit around a camp fire. Find the number of ways to
In how many ways can arrange the 6 students with an arrange them if
identical twins on a round table?
(i) tidak ada halangan.
Penyelesaian:
there is no restrictions.
Oleh sebab meja ialah bulat, cara susunan ialah
(ii) semua lelaki mesti duduk sebelah-
(6 − 1)! = 120 jika semua orang adalah berbeza. menyebelah.

Tetapi ada sepasang anak kembar yang sama, all the boys must sit side by side.
(6−1)!
maka cara susunan ialah 2! = 60. (iii) semua perempuan mesti diasingkan

Since the table is round, the number of ways to arrange all the girls must be separated.
is (6 − 1)! = 120 if all the people are different. But there
(i) Bilangan cara menyusun = (10 − 1)!
is an identical twins, hence the number of ways is Number of ways to arrange = 362 880
(6−1)! (ii) Bilangan cara menyusun
2! = 60. Number of ways to arrange
= (5 − 1)! × 6!
(b) 5 orang wanita dan 4 orang lelaki duduk dalam = 17 280
satu bulatan untuk bermain satu permainan. (iii) Bilangan cara menyusun
Cari bilangan cara menyusun semua jika Number of ways to arrange
= (10 − 1)! − (7 − 1)! × 4!
5 women and 4 men sit in a circle to play a game. = 345 600
Find the number of ways to arrange all of them if
(c) Digit-digit 1, 2, 3, 4, 5, 6, 7 dan 8 disusun dalam
(i) satu pasang suami isteri mesti duduk satu bulatan. Cari bilangan cara menyusun
bersama. jika

a couple of husband and wife must sit together. The digits 1, 2, 3, 4, 5, 6, 7 and 8 are arranged in a
circle. Find the number of ways to arrange them if
(ii) tiga orang perempuan tertentu mesti
duduk bersama. (i) semua digit ganjil mesti bersama.

three specific women must sit together. all the odd digits must be together.

(i) Bilangan cara = (8 −1)! × 2! (ii) digit nombor ganjil berselang-seli dengan
Number of ways = 10 080 digit nombor genap.
(ii) Bilangan cara menyusun = (7 − 1)! × 3!
Number of ways to arrange = 4 320 the odd digits are alternate to the even digits.

(i) Bilangan cara menyusun = (5 − 1)! × 4!
Number of ways to arrange = 576
(ii) Bilangan cara menyusun = (4 − 1)! × 4!
Number of ways to arrange = 144

90

  Matematik Tambahan  Tingkatan 5  Bab 4 Pilih Atur dan Gabungan 

4.2 Gabungan

Combination

NOTA IMBASAN BAB 4

1. Jika r objek dipilih daripada n objek tanpa 2. Perhatikan bahawa
Note that
mempertimbangkan tertib susunannya, bilangan
(i) nCn = 1
gabungan yang berlainan adalah diberi oleh (ii) nCn − 1 = n
(iii) nC0 = 1
If r objects are chosen from n objects without arranging them,the (iv) nCr = nCn − r
(v) r! nCr = nPr
number of different combinations is given by

nCr = (n n! .
− r)!r!
n(n − 2)….3.2.1
= (n − r)(n − r −1)…3.2.1.r(r − 1)(r − 2)…3.2.1

= n(n − 1)...(n – r + 1)
r(r −1)(r − 2)…3.2.1

9. Selesaikan setiap yang berikut.

Solve each of the following. TP 2

CONTOH (a) Buktikan / Prove that
(i) r! nCr = nPr
Buktikan / Prove that (ii) nC0 = 1
(i) nCn = 1
(ii) nCr = nCn − r (i) Sebelah kiri = r! nCr
Let side =
Penyelesaian: r!n!
(n − r)!r!
(i) nCn = (n n!
− n)!n!
n!
= n! =1 = (n − r)!
(0)!n!

(ii) Sebelah kiri / Left side = nPr Sebelah kanan

n! Right side
(n − r)!r!
= nCr = Maka/Hence, r! nCr = nPr

Sebelah kanan / Right side (ii) nC0 = (n n!
− 0)!0!
n!
= nCn − r = (n − n + r)!(n – r)! = n! = 1
(n)!
n!
= r!(n − r)! = sebelah kanan

Right side

Maka / Hence nCr = nCn − r

91

  Matematik Tambahan  Tingkatan 5  Bab 4 Pilih Atur dan Gabungan

(b) Nilaikan / Evaluate (c) Carikan / Find
(i) 8C2 (i) 10C7 ÷ 6C3
(ii) 4C2 × 9C4 (ii) nCn − 2

BAB 4 (i) 8C2 = (8 8! (i) 10C7 ÷ 6C3 = 10! ÷ (6 6!
– 2)!2! (10 – 7)!7! – 3)!3!

= 8! = 6
(6)!2!
(ii) nCn – 2
= 28 n!
= 2)!(n
(n − n + – 2)!
Sudut Kalkulator
n!
Tekan = (2)!(n – 2)!
8 SHIFT nCr 2 = 28
n(n – 1)(n – 2)!
= (2)!(n – 2)!

(ii) 4C2 × 9C4 = n(n – 1)
(2)
= 4! × 9!
(4 – 2)!2! (9 – 4)!4!

= 756

10. Selesaikan setiap yang berikut.

Solve each of the following. TP 2

CONTOH (a) Daripada sekumpulan 8 orang pekerja, 4

Cari bilangan cara memilih 3 orang sukarelawan orang pekerja akan dipilih secara rawak
daripada 12 orang sukarelawan.
untuk menghadiri satu seminar peningkatan
Find the number of ways to choose 3 volunteers from
12 volunteers. kemahiran. Cari bilangan cara boleh

Penyelesaian: dilakukan.

Oleh sebab pilihan dan susunan ini tidak penting, From a group of 8 workers, 4 workers are chosen
randomly to attend a skill improvement seminar.
Find the number of ways to do this.

ini adalah masalah gabungan. Bilangan cara pilihan = 8C4

Since this selection and arrangement is not important, Number of ways = 70

this is a combination problem.

Bilangan cara / Number of ways

= 12C3 = 12!
(12−3)!(3)!

= 220

(b) Seorang guru ingin memilih 5 orang pelajar (c) Di sebuah gerai makan, terdapat 7 jenis lauk-

secara rawak daripada 12 orang pelajar di pauk untuk dipilih. Awang ingin memilih

dalam kelasnya untuk menyertai suatu 4 jenis makanan, Cari bilangan cara dia boleh

pertandingan perbahasan. Cari bilangan cara memilih jika semua lauk-pauk adalah sedap.

yang boleh dilakukan . At a food stall, there are 7 types of side dishes to
choose. Awang wants to choose 4 types of food,
A teacher wants to choose 5 students randomly from Find the number of ways he can choose if all the
the class of 12 students to take part in a debate. Find side dishes are delicious.
the number of ways to do this.

Bilangan cara pilihan = 12C5 Bilangan cara pilihan = 7C4
Number of ways = 35
Number of ways = 792

92

  Matematik Tambahan  Tingkatan 5  Bab 4 Pilih Atur dan Gabungan 

11. Selesaikan setiap soalan bersyarat yang berikut. BAB 4

Solve each of the following questions with the given conditions. TP 5

CONTOH

Satu jawatan kuasa kebersihan terdiri daripada 5 orang yang dipilih daripada 5 orang guru lelaki, 6 orang

guru perempuan dan seorang pengetua dari suatu sekolah. Cari bilangan cara memilih jawatan kuasa

itu jika

A cleanliness committee is formed by 5 people chosen from 5 male teachers, 6 female teachers and a principal of a
school. Find the number of ways to select the committee if

(i) tiada sebarang syarat dikenakan

there is no given conditions.

(ii) jawatan kuasa itu mesti ada pengetua.

the committee must have principal.

(iii) jawatan kuasa itu mesti ada pengetua dan dua orang guru lelaki.

the committee must have the principal and two men teachers.

Penyelesaian:

(i) Jumlah orang = 12 dan memerlukan 5 orang.
Total people = 12 and we need 5 people only.

Maka, tbhielannugmabnecr aorfaw=ay1s2C=512=C57=92792

Hence,

(ii) Pengetua mesti ada di dalam jawatan kuasa, bilangan cara memilih pengetua = 1C1.

The principal must be in the committee, the number of ways to select the principal = 1C1.

Masih ada 4 tempat untuk diisi oleh 11 orang guru, maka bilangan cara memilih guru t=ea1c1hCe4r.s = 11C4.

There are 4 more places to be filled up by 11 available teachers, hence the number of ways to select

Dengan menggunakan petua pendaraban, jumlah bilangan cara = 11C4 × 1C1 = 330

By using multiplication rule, the total number of ways = 11C4 × 1C1 = 330

(iii) Jawatan kuasa memerlukan 1 pengetua, 2 guru lelaki dan 2 guru perempuan. Maka, jumlah bilangan

cara

The committee needs 1 principal, 2 male and 2 female teachers. Thus, the total number of ways = 1C1 × 5C2 × 6C2
= 150

(a) 6 titik telah dilukis pada suatu kertas dengan keadaan tiada 3 titik adalah segaris. Cari bilangan

6 points are marked on a paper such that there is no 3 points on a straight line. Find the number of

(i) garis lurus boleh dibina.

straight line that can be formed.

(ii) segi tiga boleh dibina.

triangles that can be formed.

(iii) segi empat boleh dibentuk.

rectangles that can be formed.

(i) Untuk melukis satu garisan, kita memerlukan mana-mana dua titik sahaja.

To draw a straight line, we neeed any two points only.

Maka, bilangan cara = 6C2 = 15.

So, number of ways

(ii) Untuk melukis satu segi tiga, kita memerlukan mana-mana tiga titik sahaja.

To draw a triangle, we need any three points only.

Maka, bilangan cara = 6C3 = 20.

So, number of ways

(iii) Untuk melukis satu segi empat, kita memerlukan mana-mana empat titik sahaja.

To draw a rectangle, we need any four points only.

Maka, bilangan cara = 6C4 = 15.

So, number of ways

93

  Matematik Tambahan  Tingkatan 5  Bab 4 Pilih Atur dan Gabungan

BAB 4 (b) Suatu pasukan tenis terdiri daripada 5 orang yang dipilih daripada 4 orang lelaki dan 5 orang

perempuan. Cari bilangan cara boleh dilakukan jika pasukan itu

A tennis team is formed by 5 people, which is chosen from 4 men and 5 women. Find the number of ways to form
the team if the team

(i) mesti mengandungi seorang lelaki sahaja.

must have only one man.

(ii) mesti mengandungi lebih perempuan daripada lelaki.

must have more women than men.

(iii) mesti mengandungi satu pasang lelaki dan perempuan yang tertentu.

must have a pair of males and a specific female.

(i) Bilangan cara = 4C1 × 5C4 = 20 Kes 3: 5 perempuan, 0 lelaki

Number of ways Bilangan cara = 5C5 = 1

(ii) Kes 1: 3 perempuan, 2 lelaki Number of ways

Bilangan cara = 5C3 × 4C2 = 60 Jumlah cara = 81

Number of ways (iii) Bilangan cara = 6C2 = 15

Kes 2: 4 perempuan, 1 lelaki Number of ways

Bilangan cara = 5C4 × 4C1 = 20

Number of ways

(c) Terdapat 8 orang ingin pergi ke bandar menggunakan dua buah kereta yang masing-masing hanya
boleh membawa 4 orang sahaja. Terdapat hanya tiga daripada 8 orang itu ada lesen memandu kereta.
Cari bilangan cara mereka boleh pergi ke bandar.

There are 8 people who want to go to town in two cars where each car can only take 4 people. There are only three
of the 8 people have car driving license. Find the number of ways of them to go to town.

Bilangan cara = 2 [3C1 × 2C1 × 6C3 × 3C3]
Number of ways = 240

(d) Suatu hadiah yang mengandungi 5 bahan akan dipilih daripada 10 pensel berwarna-warni, 7 batang
pen dan 6 batang pembaris. Cari bilangan cara untuk membungkus hadiah jika

A prize which consists of 5 items will be chosen from 10 colour pencils, 7 pens and 6 rulers. Find the number of
ways to package the prize if

(i) tiada halangan.

there is no restriction.

(ii) hadiah mesti mengandungi sekurang-kurangnya satu bahan daripada tiga kategori itu.

it must contain at least one item from the three categories.

(iii) hadiah tidak mengandungi pembaris.

it does not contain rulers.

(i) Bilangan cara = 23C5 = 33 649
Number of ways
(ii) Bilangan cara = 10C1 × 7C1 × 6C1 × 20C2
Number of ways = 79 800
(iii) Bilangan cara = 17C5 = 6 188
Number of ways

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  Matematik Tambahan  Tingkatan 5  Bab 4 Pilih Atur dan Gabungan  BAB 4

(e)

Rajah menunjukkan dua baris kerusi, satu baris berdekatan dengan pintu keluar.
Cari bilangan pilihan cara tempat duduk bagi 7 orang yang terdiri daripada 2 kanak-kanak, 1 pasang

suami-isteri dan 4 orang dewasa jika

The diagram shows two rows of chairs, one row is near to the exit door.
Find the number of ways of seats for 7 people that is formed by 2 children, 1 couple of husband-wife and 4 adults

if

(i) 2 kanak-kanak mesti duduk dalam baris yang sama.

2 children must sit in the same row.

(ii) 1 pasang suami-isteri dan 2 kanak-kanak mesti duduk pada baris yang sama.

1 couple of husband-wife and 2 children must sit in the same row.

(i) Jika memilih baris tiga kerusi.
If choose the row of three chairs
Bilangan cara = 5C1 × 4C4 = 5
Number of ways
Jika memilih baris empat kerusi

If choose the row of four chairs

Bilangan cara = 5C2 × 3C3 = 10
Jumlah = 15
(ii) 1 cara sahaja

Only 1 way

(f) 4 huruf dipilih daripada perkataan PEMBARIS. Cari bilangan cara jika pilihan itu

4 letters are chosen from the word PEMBARIS. Find the number of ways to choose the letters if it

(i) mengandungi huruf B.

contains the letter B.

(ii) tidak mengandungi vokal.

does not contain a vowel.

(iii) mengandungi huruf E atau R.

it contains letter E or R.

(i) Bilangan cara = 7C3 = 35
Number of ways
(ii) Bilangan cara = 5C4 = 5
Number of ways
(iii) Bilangan cara = 7C3 × 2 = 70
Number of ways

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  Matematik Tambahan  Tingkatan 5  Bab 4 Pilih Atur dan Gabungan 14

PRAKTIS SPM

BAB 4 Kertas 1 3. Dalam almari Nur, 3 daripada 5 helai baju ialah
1. (a) Diberi 7Cn > 1, senaraikan semua nilai yang biru, dan 2 daripada 4 pasang seluar ialah biru.
Pada suatu hari, dia memilih 1 baju dan 1 pasang
mungkin bagi n. seluar untuk temuduga . Dicadangkan bahawa
dia tidak boleh memakai warna sama bagi baju
2018 Given 7Cn > 1, list all the possible values of n. dan seluar. Cari berapa cara dia dapat memilih

(b) Diberi bahawa xCm = xCn, ungkapkan x dalam 2019 pakaiannya.
sebutan m dan n.
In Nur’s wardrobe, 3 out of 5 blouses are blue, and 2 out
Given that xCm = xCn, express x in terms of m and n. of 4 pants are blue. On a certain day, she picks 1 blouse
and 1 pair of pants for her interview. She is advised not to
(a) 1 ≤ n < 7 dengan n ialah integer wear the same colour for her blouse and pants. Find the
number of ways she can choose her clothes.
(b) xCm = xCn
Kes 1: Baju biru
(x − x! = (x x!
m)!m! − n)!n! Case 1: Blue blouses

n + m = x 3C1 × 2C1 = 6 cara
Kes 2: Seluar biru
2. Rajah menunjukkan kata laluan bagi mangga
Yong. Case 2: Blue pants

2019 The diagram shows a password for Yong’s padlock. 2C1 × 2C1 = 4 cara
Kes 3: Bukan baju biru dan seluar biru
0 242
Case 3: Not blue blouses and blue pants
Yong hendak menetapkan semula kata laluan,
2C1 × 2C1 = 4 cara
dengan keadaan kod baharu itu tidak boleh Jumlah cara = 14 cara

mengandungi digit 2 diikuti oleh digit 4. Berapa Total ways

bilangan kod laluan berbeza yang dapat dibentuk? Praktis
SPM
Yong wants to reset the password, where the new code Ekstra
cannot consist of digit 2 followed by digit 4. How many
different passwords can be formed?

4! – 3! = 6
2!

Sudut KBAT KBAT

Ekstra

Dengan menggunakan digit-digit 0, 1, 2, 4, 6 atau 7, (c) adalah nombor dengan lima digit dan berakhir
dengan nombor ‘0’.
cari bilangan nombor yang dibentuk jika nombor itu
is a five digit number and ends with a number ‘0’.
By using the digits 0, 1, 2, 4, 6 or 7, find the number of ways
that can be formed if the number (a) Kes 1: nombor berdigit 4 = 3 × 5P3 = 180
Kes 2: nombor berdigit 5 = 5 × 5P5 = 600
(a) lebih daripada 3 000 tanpa mengulangi digit. Kes 3: nombor berdigit 6 = 5 × 5P5 = 600
Jumlah = 600 + 600 + 180 = 1380
is more than 3 000 without repeating the digits. (b) 2 × 4 × 3 + 2 × 5 × 4 × 3 = 144
(c) 5P4 × 1P1 = 120
(b) di antara 2 500 dan 7 000 tanpa mengulangi

digit.

is between 2 500 and 7 000 without repeating the
digits.

Quiz 4

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BAB   Matematik Tambahan  Tingkatan 5  Bab 5 Taburan Kebarangkalian 

5 Taburan Kebarangkalian BAB 5
Probability Distribution
NOTA IMBASAN

5.1 Pemboleh Ubah Rawak

Random Variable

NOTA IMBASAN 4. Taburan kebarangkalian pemboleh ubah rawak diskret
ialah suatu taburan menunjukkan kebarangkalian
1. Pemboleh ubah rawak ialah suatu pemboleh ubah semua kesudahan yang mungkin dalam bentuk graf,
dengan nilainya ialah kesudahan numerik yang dapat jadual atau rajah pokok.
ditentukan daripada suatu fenomena rawak.
The probability distribution of a random discrete variable is a
A random variable is a variable whose value is an outcome is distribution to show the probability of all the possible outcomes
numeric from a random phenomena. in the form of graph, table or a tree diagram.

2. Pemboleh ubah rawak diskret ialah suatu pemboleh 5. Nilai setiap kebarangkalian P(X = x) berada di antara 0
ubah yang dapat dikira. Kesudahan yang mungkin, X dan 1, iaitu 0 ≤ P(X = x) ≤ 1.
dapat ditulis dalam tata tanda set, X = {x: x = 0, 1, 2, 3, ….}.
The value of each probability P(X = x) is between 0 and 1, that is
A discrete random variable is a variable which can be counted. 0 ≤ P(X = x) ≤ 1.
The possible outcomes, X is written in set notation, X = {x: x = 0, 1,
2, 3, ….}. 6. Hasil tambah kebarangkalian bagi semua kesudahan
yang mungkin ialah 1. ∑0nP[X = x] = 1, x = 0, 1, 2, . . . , n
3. Pemboleh ubah rawak selanjar ialah suatu pemboleh
ubah Y yang nilainya tidak dapat dikira dan ditulis The sum of probability of all the possible outcomes is 1.∑n0P[X = x]
dalam bentuk Y = {y : a < y ≤ b}, dengan keadaan a dan b = 1, x = 0,1, 2, . . . , n
adalah pemalar.

A continuous random variable is a variable Y whose value cannot
be counted and is written in the form Y = {y : a < y ≤ b} , where a
and b are constants.

1. Nyatakan pemboleh ubah rawak bagi setiap situasi yang berikut.

State the random variable for each of the following situations. TP 1
CONTOH

(i) Satu duit sylling yang adil dilambungkan sekali.

A fair coin is tossed once.

(ii) Jisim sebiji nanas di dalam bakul adalah di antara 0.5 kg dan 2.8 kg.

The mass of a pineapple in the basket is between 0.5 kg and 2.8 kg.

Penyelesaian:
(i) Pemboleh ubah ialah gambar di permukaan atas duit syiling, iaitu {kepala, bunga}.

The variable is the picture on the upper surface of coin, that is {head, flower}.

(ii) Pemboleh ubah ialah jisim nanas

The variable is the mass of the pineapple.

(a) Keputusan permainan catur.

The results of a chess game.

Pemboleh ubah ialah menang, tewas atau seri

The variable is win, lose or draw

97