Module & More Add Math Tg 5

BAB   Matematik Tambahan  Tingkatan 5  Bab 1 Sukatan Membulat 

1 Sukatan Membulat BAB 1
NOTA IMBASCAiNrcular Measure

1.1 Radian

Radian

NOTA IMBASAN 3. Apabila panjang lilitan ialah 2πj, maka sudut yang
tercangkum ialah 2π radian.
1. Sudut boleh diukur dalam darjah dan minit atau radian
(rad). When the circumference is 2pj, then the subtended angle is 2p
radian.
Angle can be measured in degrees and minutes or in radians
(rad). 4. Diketahui bahawa sudut bulatan ialah 360°, maka 2π
radian adalah setara dengan 360°.
2. Sudut yang dicangkum pada pusat bulatan oleh
lengkok yang sama panjang dengan jejarinya It is known that the angle of a circle is 360° , then 2π radian is
ditakrifkan sebagai 1 radian. Lihat contoh dalam rajah. equivalent to 360°.

The angle subtended at the centre of the circle by the arc length 5. Secara am, kita boleh tulis
which is the same length as the radius is defined as 1 radian. In general, we can write

Oj j q = q rad
j 1 rad O 2 rad 2j 360° 2π rad

j j

1. Tukarkan setiap yang berikut kepada radian. 

Convert each of the following to radians. TP 1

CONTOH (a) 43°32’

(i) 65°42’ (ii) 279°24’ Gunakan/Use

Penyelesaian: 43°32’ = θ rad
(i) Gunakan/Use 360° 2π rad

θ = θ rad θ = 43°32’ × 2π
360° 2π rad 360°

65°42’ θ rad = 0.76 rad
360° 2π rad
=

θ= 65°42’ × 2π = 1.147 rad
360°

(ii) θ = 279°24’ × 2π = 4.876 rad
360°

1

  Matematik Tambahan  Tingkatan 5  Bab 1 Sukatan Membulat

(b) 157°16’ (c) 247°54’

Gunakan/Use Gunakan/Use

BAB 1 157°16’ = θ rad 247°54’ = θ rad
360° 2π rad 360° 2π rad

θ = 157°16’ × 2π θ = 247°54’ × 2π
360° 360°

= 2.745 rad = 4.327 rad

(d) 303.68° (e) 354.74°

Gunakan/Use Gunakan/Use

303.68° = θ rad 354.74° = θ rad
360° 2π rad 360° 2π rad

θ = 303.68° × 2π θ = 354.74° × 2π
360° 360°

= 5.3 rad = 6.191 rad

2. Tukarkan setiap yang berikut kepada darjah dan minit.

Convert each of the following into degree and minutes. TP 1

CONTOH (a) 2 π rad
3
3.2 rad
2 π rad
Penyelesaian: θ° 3
360° =
Gunakan/Use 2π rad

θ = θ rad 360° × 2 π
360° 2π rad 2π 3
θ° =
3.2 rad
θ = 2π rad = 120°
360°

θ = 360° × 3.2
2π

= 183.21’ atau/or 183.35°

(b) 4 rad (c) 4.65 rad
9
4 θ° 4.65 rad
θ° 9 rad 360° = 2π rad
360°
= 2π rad 360° × 4.65
2π
360° × 4 θ° =
9
θ° = = 266.43° atau/or 266°26’
2π

= 25.46° atau/or 25°28’

(d) 1.67π rad (e) 0.72π rad

θ° = 1.67π rad θ° = 0.72π rad
360° 2π rad 360° 2π rad

θ° = 360° × 1.67π θ° = 360° × 0.72π
2π 2π

= 300.6° atau/or 300°36’ = 129.6° atau/or 129°36’

2

  Matematik Tambahan  Tingkatan 5  Bab 1 Sukatan Membulat 

1.2 Panjang Lengkok Suatu Bulatan
NOTAArcILMeBngAthSoAfNa Circle

NOTA IMBASAN BAB 1

1. Panjang lengkok, s, suatu bulatan berkadaran dengan 5. Panjang lengkok major adalah lebih panjang daripada
sudut yang tercangkum di pusat bulatan. Maka, panjang lengkok minor.
semakin besar sudut yang dicangkum, semakin besar
panjang lengkok. The major arc length is longer than the minor arc length.

The arc length, s, of a circle is directly proportional to the size of 6. Kawasan berlorek yang dibatasi oleh lengkok AB dan
the angle subtended at the centre of the circle. Hence, the bigger perentas AB dinamakan tembereng bulatan.
the angle, the longer the arc length.
The region bounded by the arc length AB and the chord AB is
2. Secara am, kita boleh tulis seperti yang berikut. called a segment.
In general, we can write as follows.
A
q° = q rad = panjang lengkok (arc length) j
360° 2π rad panjang lilitan (circumference) Oθ

dengan panjang lilitan = 2πj dan jejari ialah j. B

where the circumference = 2πj and the radius is j.

3. Pasangan yang digunakan untuk menyelesaikan 7. Perentas AB dapat diperolehi dengan petua kosinus,
masalah bergantung kepada sudut yang diberi dalam
darjah atau radian. AB = j2 + j2 − 2j2 kosθ dengan θ dalam darjah atau

The pair to solve a problem will depend on whether the angle AB = 2j sin θ
given is in degree or radian. 2

The chord AB can be obtained by using cosine rule,

4. Jika sudut θ ialah dalam radian, panjang lengkok boleh AB = j2 + j2 − 2j2 cosθ with the angle θ is in degree or
diberi oleh s = jθ.
AB = 2j sin θ .
If the angle θ is in radian, the arc length can be given by s = jθ. 2

3. Tentukan panjang lengkok, s bagi setiap bulatan yang diberi. 

Determine the arc length, s for each of the following given circles. TP 1

CONTOH (a) As B

Penyelesaian: 7.4 cm 0.8 rad
O
Oleh kerana sudut ialah O 3.2 cm A
1.2 rad
dalam radian, kita pilih s

Since the angle is in radian, we
choose
panjang lengkok, s/ arc length, s B
θ rad = 2πj panjang lengkok, AB
2π rad
1.2 rad
2π rad = panjang lengkok, s/ arc length, s 0.8 rad = Arc length AB
2πj 2π rad
2πj

Panjang lengkok, s Panjang lengkok, s
Arc length, s
Arc length, s

= 1.2 rad × 2π(3.2) Kaedah Alternatif = 0.8 rad × 2π(7.4)
2π rad 2π rad
Guna/Use AB = jθ
= 3.2(1.2)
= 3.84 cm = 3.84 cm = 5.92 cm

3

  Matematik Tambahan  Tingkatan 5  Bab 1 Sukatan Membulat (c)

(b) AB
125Њ
BAB 1 O 4.5 cm O 10.2 cm
0.5␲ rad B
s
As

0.5π rad = panjang lengkok, s Sudut yang dicangkum = 360° – 125° = 235°
2π rad 2πj
Angle subtended

Panjang lengkok, s θ = panjang lengkok, s
Arc length, s 360° 2πj

= 0.5π rad × 2π(4.5) Panjang lengkok, s
2π rad
Arc length, s

= 7.07 cm = 235° × 2π(10.2)
360°

= 41.84 cm

(d) (e)

s s

O 20.5 cm A 8.4 cm B

1.3 rad O
AB
4.5 rad

Sudut yang dicangkum = (2π − 1.3) rad Sudut yang dicangkum = (2π − 4.5) rad
Angle subtended
Angle subtended

(2π – 1.3) rad = panjang lengkok, s (2π – 4.5) rad = panjang lengkok, s
2π rad 2πj 2π rad 2πj

Panjang lengkok, s Panjang lengkok, s
Arc length, s Arc length, s

= (2π – 1.3) rad × 2π(20.5) = (2π – 4.5) rad × 2π(8.4)
2π rad 2π rad

= 102.16 cm = 15 cm

4

  Matematik Tambahan  Tingkatan 5  Bab 1 Sukatan Membulat 

4. Tentukan jejari bulatan, j dengan diberikan panjang lengkok, s dan sudut bagi setiap bulatan yang berikut.

Determine the radius, j of the given the arc length, s and the angle of each of the following circles. TP 2

CONTOH (a) BAB 1

20.3 cm

8.4 cm 204Њ
AB O

1.4 rad j
j

O

Penyelesaian: 204° = panjang lengkok, s
360° 2πj

Oleh kerana sudut ialah dalam radian, kita pilih j = 20.3 × 360°
204° × 2π
Since the angle is in radian, we choose

θ rad = panjang lengkok, s / arc length, s = 5.7 cm
2π rad 2πj

Apabila diringkaskan, kita dapat

When simplified, we get
s
j = θ

= 8.4 = 6 cm
1.4

(b) 14.5 cm (c)

Oj 0.8␲ rad
1.6 rad O

j

11.5 cm

j = 14.5 j = 11.5
2π – 1.6 2π – 0.8π

= 3.1 cm = 3.05 cm

(d) 14.2 cm (e)

Oj 3 ␲ rad
105Њ 4

Oj

3 cm

360° – 105° = panjang lengkok, s j = 3 3
360° 2πj – 4
2π π
14.2 × 360°
j = 255° × 2π = 0.76 cm

= 3.19 cm

5

  Matematik Tambahan  Tingkatan 5  Bab 1 Sukatan Membulat

BAB 1 5. Tentukan sudut tercangkum, θ dalam radian, di pusat bulatan dengan diberikan jejari bulatan, j dan panjang

lengkok, s bagi setiap bulatan yang berikut.

Determine the angle subtended, θ in radian, at the centre of the circle given the radius, j and the arc length, s for each of the

following circles. TP 3

CONTOH (a)

7.2 cm

P 4.4 cm ␪ 6.1 cm O
O ␪

9.8 cm Q 7.2
6.1
(2π – θ) =

Penyelesaian: θ = 5.1 rad

θ rad = panjang lengkok, s / arc length, s
2π rad 2πj

Ini boleh dringkaskan kepada

This can be simplified into
s
θ = j

( 2π – θ) = 9.8
4.4

θ = 4.06 rad

(b) 4.8 cm (c)

O 3.2 cm = ␪ =
␪
O
1.8 cm

2.8 cm

( 2π – θ) = 4.8 (2π – 2θ) = 2.8
3.2 1.8

θ = 4.78 rad θ = 2.36 rad

(d) (e) 11.4 cm

= ␪
␪ O

O 2.5 cm

=
=
3.2 cm
6.3 cm 8.2 cm

(2π – θ) = 3.2 2πj = 6.3 + 8.2 + 11.4
3 2.5
j = 4.12 cm

θ = 2.44 rad θ = 11.4
4.12

= 2.77 rad

6

  Matematik Tambahan  Tingkatan 5  Bab 1 Sukatan Membulat 

6. Tentukan perimeter tembereng bagi setiap bulatan berpusat O yang berikut

Determine the perimeter of the segment of each of the following circles with centre O. TP 4

CONTOH (a) BAB 1

O 6.2 cm P P
2 rad 9.6 cm

O
Q 0.4␲ rad

Q

Penyelesaian: 0.4π rad = 0.4π × 360°
2π
Perentas PQ dapat diperoleh dengan petua
= 72°
kosinus, iaitu PQ = j2 + j2 − 2j2 kos θ , dengan θ
dalam darjah. Maka/Hence,

The length of the chord PQ can be obtained by using the PQ = 9.62 + 9.62 − 2(9.6)2 kos72°
cosine rule, that is PQ = j2 + j2 − 2j2 cos θ , such that θ
is in degree. = 11.28 cm

Panjang lengkok PQ = jθ

2 × 360° Arc length PQ = 9.6(0.4π)
2π
2 rad = = 114.59° = 12.06 cm

Maka/Hence, Perimeter = (12.06 + 11.28) cm

= 23.34 cm

PQ = 6.22 + 6.22 − 2(6.2)2 kos 114.59°
= 10.43 cm

Panjang lengkok PQ = jθ = 6.2(2)
Arc length PQ

= 12.4 cm

Perimeter = (12.4 + 10.43) cm = 22.83 cm

(b) (c)

P P 3 cm 4.5 rad
8 cm O

O Q

30 cm Q

Panjang lengkok minor PQ = 2π(8) – 30 Panjang lengkok minor PQ = 2π(3) – 4.5(3)

Minor arc length PQ = 20.27 cm Minor arc length PQ = 5.35 cm

∠POQ = 20.27 ∠POQ = 2π − 4.5
8
= 1.78 rad
= 2.53 rad
= 102.17°
= 145.14°
Maka, PQ = 32 + 32 − 2(3)2 kos102.17°
Maka, PQ = 82 + 82−2(8)2 kos145.14°
= 15.27 cm Hence = 4.67 cm

Perimeter = 15.27 + 20.27 Perimeter = 5.35 cm + 4.67 cm

Hence = 35.54 cm = 10.02 cm

7

  Matematik Tambahan  Tingkatan 5  Bab 1 Sukatan Membulat

(d) (e) Q R

=

P =
P
BAB 1 Q O 5.3 cm

O ␲ rad
12
1.4␲ rad

7.3 cm

∠POQ = 150° Panjang lengkok PQR = 2π(5.3) − 5.3(1.4π)

Panjang jejari/Arc length, j Arc length PQR = 10 cm

= (7.3 × 360°) = 1.99 cm ∠POQ = (2π − 1.4π) ÷ 2
(210 × 2π)
= 0.3π

Maka/Hence, = 54°

PQ = 1.992 + 1.992 − 2(1.99)2 kos150° Panjang perentas PQ
= 3.84 cm
Length of chord PQ

Panjang lengkok = 150° × 2π(1.99) = 5.32 + 5.32 − 2(5.3)2 kos 54°
360° = 4.81 cm
Arc length
Perimeter = (4.81 × 2 + 10) cm
  = 5.2 cm
= 19.62 cm
Perimeter = 5.2 cm + 3.84 cm

= 9.04 cm

7. Selesaikan masalah yang melibatkan panjang lengkok.

Solve the problems involving the arc lengths. TP 4

CONTOH (a)

A

P 4.7 cm 9.2 cm
Q

AB O 0.7 rad DB

O

Rajah di atas menunjukkan dua bulatan berpusat Rajah di atas menunjukkan sebuah sektor

O dan masing-masing mempunyai jejari 8 cm dan AOB berpusat O dan mempunyai jejari 9.2 cm.

14 cm. Cari perimeter rantau berlorek itu. Cari perimeter rantau berlorek itu.

The diagram above shows two circles with centre O The diagram above shows a sector AOB with centre
and with radii 8 cm and 14 cm respectively. Find the O and radius 9.2 cm. Find the perimeter of the
perimeter of the shaded region. shaded region.

0.7 rad = 0.7 × 360° = 40.1°
2π

Penyelesaian: sin 40.1° = AD
9.2
∠POQ = s = 4.7
j 14 AD = 5.93 cm

= 0.336 rad OD = 9.22 − 5.932
= 7.03cm
Panjang lengkok AB
Panjang lengkok = 9.2 × 0.7
Arc length AB
Arc length = 6.44 cm
= 8(0.336) = 2.69 cm
Perimeter = 5.93 + 6.44 + (9.2 – 7.03)
Perimeter = 2.69 + 6(2) + 4.7
= 14.54 cm
= 19.39 cm

8

  Matematik Tambahan  Tingkatan 5  Bab 1 Sukatan Membulat 

(b) (c)

O MP A
␪
R O␣ P␪ BAB 1
␣ Q B

N

Rajah di atas menunjukkan dua sektor, POQ Rajah di atas menunjukkan dua sektor, AOB

dan MON. M ialah titik tengah OP. dan APB. Diberi nisbah jejari OA kepada PA

Ungkapkan α dalam sebutan θ jika perimeter ialah 3 : 1, AP = 2 cm dan perimeter rantau

MPQR dan ORN adalah sama. berlorek ialah 10 cm.

The diagram above shows two sectors, POQ and The diagram above shows two sectors, AOB and
MON. M is the midpoint of OP. Express α in terms of APB. Given that the ratio of radius OA to PA is 3 : 1,
θ if the perimeter of MPQR and ORN are the same. AP = 2 cm and the perimeter of the shaded region is
10 cm.
Katakan OM = j cm
Let (i) Tunjukkan bahawa θ = (5 − 3a) rad.

Maka, jθ + 2j + 2jθ = 2j + jα Show that θ = (5 − 3a) rad.

Hence, α = 3θ (ii) Jika a = 30°, cari nilai θ, dalam radian.

If a = 30°, find the value of θ, in radian.

(i) Diberi AP = 2 cm, maka OA = 6 cm

Given hence

6a + 2θ = 10

θ = 5 − 3a

(ii) a = 30°

= π rad
6
π
θ = 5 − 3 6 

= 3.43 rad

(d) A (e) K

␲ O- M
3
P C ␲

4 =
=

B O

NL

Rajah di atas menunjukkan satu semibulatan Rajah di atas menunjukkan sektor KOL

berpusat O dan satu sektor berpusat P dan berpusat O dan MLN berpusat L. Diberi bahawa

∠APB = π . Cari jejari PA jika perimeter rantau OM = ML = 5 cm, dan ∠MON = π rad, cari
3 4
35π perimeter rantau berlorek itu.
berlorek ialah 6 cm.
The diagram above shows sector KOL with centre O
The diagram above shows a semicircle with centre O
and sector MLN with centre L. Given that OM = ML =
π
and a sector with centre P and ∠APB = 3 . Find the 5 cm, and ∠MON = π rad, find the perimeter of the
4
radius PA if the perimeter of the shaded region is shaded region.

35π cm. Perimeter = OM + ON + panjang lengkok/arc
6
length MN + MK + panjang lengkok/

π = 60° OA = j arc length KL + ML
3 2
= 5 + (5 2 − 5) + 5 π  + (5 2 − 5) +
Maka, AP = AB = j. 35π πj πj 5 4
Hence = 6 = 3 + 2  π +5
2 4

j = 7 cm = 23.62 cm

9

  Matematik Tambahan  Tingkatan 5  Bab 1 Sukatan Membulat

NOTA IMBASAN

1.3 Luas Sektor Suatu Bulatan

Area of Sector of a Circle

BAB 1 NOTA IMBASAN

1. Didapati bahawa luas sektor AOB bagi suatu bulatan 3. Luas tembereng/Area of a segment A
adalah berkadaran dengan saiz sudut, θ rad yang
tercangkum di pusat. = luas sektor – luas segi tiga Oθ
j
It is known that the area of the sector AOB of a circle is area of sector − area of triangle
proportional to the size of the angle, θ rad which is subtended at
the centre. = θ j2 – 1 j2 sin θ
2 2

B

Maka, luas sektor 4. Dari apa yang telah dipelajari, hubungan secara am
antara sudut, panjang lengkok dan luas sektor adalah
Hence, the area of a sector Oj seperti berikut.
θ
= θ × πj2 A From the earlier studies, the general relationship between the
2π B angles, arc length and the area of sector are as follows.

= θ j2
2

2. Jika diungkapkan dalam nisbah, kita dapat 3q60° ° = q rad = panjang lengkok, s = luas sektor
2π rad 2πj luas bulatan
luas sektor = θ = q°
luas bulatan 2π 360° θ° = θ rad = arc length, s = area of sector
360° 2π rad 2πj area of circle
If expressed in the ratio form, we get area of sector = θ = θ°
area of circle 2π 360°

8. Tentukan luas sektor bagi setiap bulatan yang berikut. Berikan jawapan anda kepada dua tempat perpuluhan.

Determine the area of the sector for each of the following circles. Give your answer to two decimal places. TP 3

CONTOH (a)

P

3.5 cm

O 6 cm Q O 5 ␲ rad
2.3 rad 6

Q

P 5
6
Sudut yang tercangkum = π rad

Angle subtended

Penyelesaian: 5 π
6
Sudut yang tercangkum = (2π – 2.3) rad Maka, luas = (3.5)2 = 16.04 cm2
2
Angle subtended Hence, the area

Maka, luas = 2π – 2.3(6)2 = 71.70 cm2
2
Hence, the area

10

(b)   Matematik Tambahan  Tingkatan 5  Bab 1 Sukatan Membulat  BAB 1

AO B (c)
C
P

215Њ O
7.8 cm Q

AOB ialah diameter dan panjangnya ialah Sudut yang tercangkum = (360 – 215)°
10.4 cm, ∠AOC = 2∠BOC.
Angle subtended = 145°
AOB is the diameter and the length is 10.4 cm,
∠AOC = 2∠BOC. Maka, luas = 145° × π × ( 7.8)2
360°
2 Hence, the area
Sudut yang tercangkum = 3 rad
π = 76.98 cm2
Angle subtended


  2 π
3
Maka, luas = ( 5.2)2
2
Hence, the area

= 28.32 cm2

(d) (e)

6 cm 10 cm A

A 3 ␲ rad
2.5 cm 5

OB BO

Sudut yang tercangkum = 6 = 2.4 rad Jejari sektor = 10
2.5 3π
Angle subtended Radius of the sector
5
2.4
Maka, luas = 2 × (2.5)2 = 7.50 cm2 = 50 cm
3π
Hence, the area

 Maka, luas = 3 π  50 2
5 3π
Hence, the area
2

= 26.53 cm2

11

  Matematik Tambahan  Tingkatan 5  Bab 1 Sukatan Membulat

9. Tentukan jejari bagi sektor bulatan yang berikut. Beri jawapan kepada dua tempat perpuluhan.

Determine the radius of the following sectors of circles. Give the answers to two decimal places. TP 4

BAB 1 CONTOH (a)

1.6 rad

3.5 rad

Diberi luas sektor ialah 20.5 cm2. Diberi luas sektor ialah 10.21 cm2.

Given the area of the sector is 20.5 cm2. Given the area of the sector is 10.21 cm2.

Penyelesaian: Gunakan /Use

Gunakan luas sektor = θ luas sektor = θ
luas bulatan 2π luas bulatan 2p

Use area of sector = θ area of sector = θ
area of circle 2p area of circle 2p

20.5 = 1.6 10.21 = 2π – 3.5
πj2 2π πj2 2π

j2 = 20.5 × 2 j2 = 10.21 × 2
1.6 2p – 3.5

j = 5.06 cm j = 2.71 cm

(b) (c)

O 38 cm
P
P
O

12 cm Q
Q

Diberi luas sektor ialah 42 cm2 dan panjang Diberi luas sektor ialah 85 cm2 dan panjang
lengkok PQ ialah 12 cm. lengkok PQ ialah 38 cm.

Given that the area of the sector is 42 cm2 and the Given that the area of the sector is 85 cm2 and the arc
arc length of PQ is 12 cm. length of PQ is 38 cm.

Gunakan/Use Gunakan/Use

panjang lengkok = luas sektor 38 = πj2 – 85
2πj luas bulatan 2πj πj2

arc length = area of sector j = 2(πj2 – 85)
2πj area of circle 38

21π2j = 42 2(πj2 – 85) = 38j
πj2 πj2 – 19j – 85 = 0

j = 42 × 2 j = 19 ± (–19)2 – 4π(–85)
12 2π

= 7 cm j = 9.04 cm

12

(d)   Matematik Tambahan  Tingkatan 5  Bab 1 Sukatan Membulat 

(e)

88Њ O BAB 1

Diberi luas sektor ialah 7.5 cm2. 12.5 cm

Given the area of the sector is 7.5 cm2. Diberi luas sektor ialah 30 cm2.

Given the area of the sector is 30 cm2.

Gunakan/ Use Gunakan/ Use

luas sektor/area of sector = θ 30 = 12.5
luas bulatan/area of circle 360° πj2 2πj

7.5 = 88 j = 30 × 2
πj2 360° 12.5

j2 = 7.5 × 360° = 4.8 cm
88π

j = 3.13 cm

10. Tentukan sudut tercangkum, θ, dalam radian di pusat bulatan bagi yang berikut. Berikan jawapan anda kepada

dua tempat perpuluhan.

Determine the angle subtended, θ, in radian at the centre of the circle for each of the following. Give your answer to two

decimal places. TP 4

CONTOH (a)

4.8 cm
␪O

O 9.3 cm
␪

Diberi luas sektor ialah 18 cm2 Diberi luas sektor ialah 33 cm2.

Given the area of the sector is 18 cm2. Given the area of the sector is 33 cm2.

Gunakan/ Use

Penyelesaian: luas sektor = θ
luas bulatan 2p
luas sektor θ
Gunakan luas bulatan = 2π area of sector = θ
area of circle 2p
area of sector θ
Use area of circle = 2p 33 = 2π – θ
π(4.8)2 2π
18 θ
π(9.3)2 = 2π 66π = 2π – θ
23.04π

θ = 18 × 2 2.865 = 2π – θ
(9.3)2 θ = 2p − 2.865
= 3.42 radian
= 0.42 radian

13

  Matematik Tambahan  Tingkatan 5  Bab 1 Sukatan Membulat (c)

BAB 1 (b)

12 cm

j␪

Diberi luas sektor ialah 46 cm2. Diberi luas sektor ialah 30 cm2 dan perimeter
ialah 15 cm. Ungkapkan θ dalam sebutan
Given the area of the sector is 46 cm2. jejari, j.

Gunakan/Use Given the area of the sector is 30 cm2 and the
perimeter is 15 cm. Express θ in terms of the radius, j.
luas sektor = panjang lengkok
luas bulatan 2πj
2j + jθ  = 15 …… ➀
area of sector arc length
area of circle = 2πj 1 j2θ  = 30 …… ➁
2
46 12
πj2 = 2πj ➁÷➀

j = 46 × 2 = 7.67 cm 1 j2q
12 2
=2
2j + jq
s = jθ
jq
θ = 12 = 1.56 radian 2(2 + q) = 2
7.67

jθ = 8 + 4θ

θ= 8
j–4

(d) (e)

21 cm O␪

␪
8 cm

Diberi luas sektor ialah 43 cm2. Diberi luas sektor ialah 54 cm2.

Given the area of the sector is 43 cm2. Given the area of the sector is 54 cm2.

  1 × 82 × θ  = 43 Gunakan/Use
2
luas sektor panjang lengkok θ
θ = 1.34 radian luas bulatan = 2πj = 2π

area of sector = arc length = θ
area of circle 2πj 2π

54 = 21 = θ
πj2 2πj 2π

j = 54 × 2 = 5.14 cm
21

2p – θ = 21
5.14

θ = 2.2 radian

14

  Matematik Tambahan  Tingkatan 5  Bab 1 Sukatan Membulat 

11. Tentukan luas tembereng yang berlorek bagi suatu bulatan yang berikut. Berikan jawapan anda kepada dua

tempat perpuluhan.

Determine the area of the shaded segment for each of the following circles. Give your answer to two decimal places. TP 4

CONTOH (a) BAB 1

P Q

O 5.3 cm Q 8.1 cm O
2.3 rad
4 ␲ rad
3

P

Penyelesaian: Luas berlorek/Shaded area

Luas berlorek = luas sektor POQ – luas segi tiga POQ = luas sektor POQ – luas segi tiga POQ
Shaded area = area of sector POQ − area of triangle POQ
area of sector POQ – area of triangle POQ
Luas sektor/Area of sector
Luas sektor = 1 4π
= 1 j2θ = 1 (5.3)2(2.3) = 32.3 cm2 2 j2 2π − 3
2 2  Area of sector
 2 = 1 2π
2.3 rad  = 2.3 × 360° = 131.78° 2 (8.1)2 3
2π
= 68.71 cm2
Luas segi tiga/Area of triangle
2π rad = 120°
= 1 ×(5.3)2 × sin131.78° = 10.47 cm2 3
2 1
Luas segi tiga = 2 × (8.1)2 × sin 120°
Maka, luas tembereng Area of triangle
= 28.41 cm2
Hence, the area of the segment
Maka, luas tembereng = 68.71 − 28.41
= 32.3 − 10.47 Hence, the area of segment = 40.3 cm2

= 21.83 cm2

(b) Q (c)
P
O 4.3 cm
1.4 rad
B
O A

3.5 cm R 50Њ
S
5.5 cm P

Diberi POR dan SOQ adalah garis lurus yang Perentas AB = 5.52 + 5.52 − 2(5.5)2 kos50°
Chord AB = 4.65 cm
melalui pusat O.

Given that POR and SOQ are straight lines passing
through the centre O.

Luas sektor POS = luas sektor QOR Kos/ cos ∠AOB = 4.32 + 4.32 − 4.652
2(4.3)2
area of sector POS = area of sector QOR
∠AOB = 65.46°
1 j2θ = 1 (3.5)2(π − 1.4) = 1.14 rad
2 2 50° = 0.87 rad

= 10.67 cm2 AL ureaasotfesmegbmereennt g = s1212in((5540..°35))+22(s021in.8(6745)..34−)62(°211.(154.)5−)2

1.4 rad = 1.4 × 360°
2π
= 80.21°
1 × (3.5)2 × sin80.21°
Luas segi tiga = 2

Area of triangle
= 6.04 cm2
= 3.7 cm2
Maka, luas tembereng = 2(10.67 − 6.04)

Hence, the area of segment = 9.26 cm2

15

  Matematik Tambahan  Tingkatan 5  Bab 1 Sukatan Membulat

1.4 Aplikasi Sukatan Membulat

Application of Circular Measures

BAB 1 12. Selesaikan masalah yang melibatkan sukatan membulat yang berikut.

Solve the following problems involving the circular measures. TP 5

CONTOH (a)

O R
P␪ O

Q

PSQ

Rajah di atas menunjukkan sekeping kek yang Rajah di atas menunjukkan satu corak yang

telah dipotong kepada 8 keping yang bersaiz terdiri daripada satu sektor major OPRQ

sama daripada satu kek yang bulat berpusat O dan berpusat O dan sebuah semibulatan POQ

berjejari 18 cm. Cari berpusat S dengan jejari 8 cm.

The diagram above shows a piece of cake that is cut into Tunjukkan bahawa luas rantau berlorek ialah
8 pieces of the same size of a round cake at centre O and
with a radius of 18 cm. Find 32(π + 2) cm2.

(i) sudut θ dalam radian. The diagram above shows a pattern which is made
up of a major sector OPRQ with centre O and a
the angle θ in radian. semicircle POQ of centre S with a radius of 8 cm.

(ii) panjang lengkok PQ. Show that the shaded area is 32(π + 2) cm2.

the arc length PQ. OP = 8 2 cm

(iii) jumlah luas permukaan kepingan kek itu jika Luas sektor major = 270° × π(8 2)2
360°
tebal kek ialah 5 cm. Area of major sector

the total surface area of the cake if the thickness is = 96π cm2
5 cm.

Penyelesaian:

(i) 8θ = 2p. Maka/Hence, θ = 1 p Luas segi tiga = 1 × (8 2)2
4 2
Area of triangle

(ii) Panjang lengkok PQ = jθ = 64 cm2
Arc length PQ
Jumlah luas = 96π + 64
1
= 18 4 p = 14.14 cm Total area

Luas semibulatan = π(8)2 = 64π

(iii) Luas sektor/Area of sector Area of semiclrcle

= 1 × 182 × 1 p Luas rantau berlorek = 96π + 64 − 64π
2 4
Area of shaded region = 32π + 64

= 127.23 cm2 = 32(π + 2) cm2

Luas permukaan/Total surface area

= 2(18 × 5) + 14.14 × 5 + 2(127.23)

= 505.16 cm2

16

  Matematik Tambahan  Tingkatan 5  Bab 1 Sukatan Membulat 

(b) (c)

Q

Q P 10 cm S

P T R BAB 1
5 cm 8 cm
5 cm
O

10 cm 1.5 rad
O

Rajah di atas menunjukkan sebuah kipas yang Rajah di atas menunjukkan dua sayap yang

dibina daripada dua keping kain, P dan Q yang serupa bagi sebuah kipas. OPQ, OTR dan QSR

berlainan. Cari beza luas kain P dan Q. adalah tiga semibulatan dengan diameter yang

The diagram above shows a fan made up ditunjukkan. Cari
of two different pieces of clothes, P and Q.
Find the difference of the area of the clothes P The diagram above shows two similar wings of a
and Q. fan. OPQ, OTR and QSR are three semicircles with
the diameters as shown. Find
Luas kain Q
(i) perimeter keseluruhan dalam sebutan π.
Area of cloth Q
the total perimeter of the wings in terms of π.
= 1 × (20)2 × 1.5 − 1 × (15)2 × 1.5
2 2 (ii) luas keseluruhan dalam sebutan π.

the total area in terms of π.

= 300 cm2 – 168.75 cm2 (i) Lilitan/ Circumference OPQ = 9π
Lilitan/ Circumference QSR = 5π
= 131.25 cm2 Lilitan/ Circumference OTR = 4π
Perimeter = 2(9π + 5π + 4π) = 36π cm
Luas kain P

Area of cloth P

= 1 × (15)2 × 1.5 − 1 × (10)2 × 1.5 1
2 2 2
(ii) Luas semibulatan OPQ = × (9)2 × π
= 168.75 cm2 – 75 cm2
Area of semicircle OPQ
= 93.75 cm2
Luas semibulatan QSR = 1 × (5)2 × π
Beza luas 2
Area of semicircle QSR
Difference of area
1
= 131.25 cm2 − 93.75 cm2 Luas semibulatan OTR = 2 × (4)2 × π

= 37.5 cm2 Area of semicircle OTR

Luas keseluruhan = 2× 1 π × (81 + 25 – 16)
area = 90π 2
Total
cm2

17

  Matematik Tambahan  Tingkatan 5  Bab 1 Sukatan Membulat SPM 11

PRAKTIS

BAB 1 Kertas 1 (a) sudut θ, dalam radian,

1. Rajah menunjukkan pandangan hadapan the angle θ, in radian,
sebahagian mural berbentuk segi empat tepat
(b) luas rantau berlorek itu.
2014 10 m × 8 m. ABP dan DCP adalah dua sektor
serupa berpusat P dan BC adalah selari dengan FE. the area of the shaded region.
Bahagian berlorek perlu dicatkan.
( a) 82 = 152 + 152 – 2(15)2 kos θ
The diagram shows a front view of a rectangular mural kos θ =
of 10 m × 8 m. ABP and DCP are two similar sectors 152 + 152 – 82
with centre P and BC is parallel to FE. The shaded region 2(15)(15)
needs to be painted.
θ = 30.93°

30.93° × 2π = 0.54 rad
360°
F 10 m E

B 4m C (b) Luas tembereng

8m Area of segments

= 1 (15)2(0.54)− 1 (15)2 sin30.93°
2 2

AP D = 2.93 cm2

Cari luas bahagian berlorek itu.

Find the area of the shaded region.

42 = 52 + 52 – 2(5)(5) kos θ 3. Rajah menunjukkan satu bulatan dengan pusat O
dan jejari 7 cm. Diberi bahawa panjang lengkok
kos θ = 52 + 52 – 42 F 10 m E
2(5)(5) 2016 minor AB ialah 10 cm, cari

θ = 47.16° B 4m C 8m The diagram shows a circle with centre O and radius
∠APB = ∠CPD 7 cm. Given that the minor arc length AB is 10 cm, find

= 180° − 47.16° ␪
2 A 5m P
D
= 66.42°
Luas sektor APB = 66.42° × π(5)2 = 14.49 cm2 ␪
360° OB
Area of sector APB
1
Luas ∆BCP = 2 (5)(5)sin 47.16°

= 9.17 cm2 A

Luas rantau berlorek = 10 × 8 − 9.17 – 2(14.49) (a) sudut θ, dalam radian,
Area of shaded region = 41.85 m2
the angle θ, in radian,
2. Rajah menunjukkan sebuah sektor berpusat O.
(b) luas sektor minor.
The diagram shows a sector with centre O.
the area of the minor sector.
2015
(a) 10 = 7(2π − θ)
P 7θ = 14p − 10
Q θ = 4.85 rad

15 cm  (b) Luas = 1 j2θ = 1 (7)2(2p − θ)
2 2
␪ Area
O = 1 10
2 (7)2 7

Diberi bahawa panjang perentas ialah 8 cm, cari = 35 cm2

Given the length of the chord is 8 cm, find

18

  Matematik Tambahan  Tingkatan 5  Bab 1 Sukatan Membulat 

Kertas 2 (a) Hitung nilai minimum, dalam cm, bagi BAB 1

1. Rajah menunjukkan sebuah semibulatan ABC panjang dan lebar kad kepada integer yang
berpusat O dan berjejari 12 cm. BAD ialah satu sektor
terdekat.
2014 berpusat A dan BE adalah berserenjang dengan AC
dengan keadaan E adalah titik tengah OC. Calculate the minimum value, in cm, of length and
breadth, of the card to the nearest integer.
The diagram shows a semicircle with centre O and
radius 12 cm. BAD is a sector with centre A and BE is (b) Seterusnya, cari luas bagi kad yang tidak
perpendicular to AC such that E is the midpoint of OC.
digunakan.
B
Hence, find the area of the unused card.

(a) PQ2 = 152 + 52 7.5 cm
P
= 250
15 cm
A O EDC PQ = 5 10 cm
Q
Cari/Find Panjang lengkok AB = 2π(7.5) 12.5 cm
(a) sudut BOD dalam radian,
Arc length AB = 15π
the angle of BOD in radian,
Panjang lengkok CD = 2π(12.5)
(b) panjang lengkok BD,
Arc length CD = 25π
the arc length BD,
Maka, 15π = jq ….. ➀
(c) luas berlorek rantau dalam cm2.
25π = (j + 5 10 )q ….. ➁ E Oj B D
the shaded area in cm2. ➁÷➀ ␪ 5√10

( a) kcoossθθ = 6 = 1 53 = j+5 10 A
12 2 j
C
π B 2j = 15 10
θ = 60° = 3 rad ␪
2␪
(b) kos 30° = 18 A 12 cm O E D C j = 15 10 cm
cos 30° AB 2
6 cm 15π × 2
15 10
AB = 20.78 cm q = = 1.99 rad

π
6
 
Panjang lengkok BD = 20.78 = 113.84°

Arc length BD kos ∠EOC = BO
= 10.88 cm OC

(c) 1 π
2 6
 
Luas sektor BAD = (20.78)2  15 10 +5 10  kos(180° − 113.84°) = EO
2
Area of sector BAD
= 113.05 cm2
1 EO = 15.98 cm
Luas ∆ABE = 2 (18)(20.78) sin30°
Area Panjang kad = 15.98 + 15 10 +5
= 93.51 cm2 2 10
Length of card

Luas rantau berlorek = 19.54 cm2 = 55.5 cm

Area of shaded region ≈ 56 cm

2. Rajah (a) menunjukkan topi lampu dan rajah (b) Lebar kad / Breadth of card
= j + 5 10
menunjukkan bentangannya dalam bentuk dua
2015 sektor berpusat O yang dilukis pada sekeping kad = 15 10 +5 10
2
berbentuk segi empat tepat.
= 39.53 cm
Diagram (a) shows the cover of a lamp and diagram
≈ 40 cm
(b) shows the net of the cover drawn on a piece of a
(b) Luas kad tidak digunakan
rectangular cardboard.
Area of unused card
15 cm O
39.532
= 56 × 40 −  2 × 1.99 +

15 cm 21  15 10 2(1.99)
2

25 cm = 1244.88 cm2
(a)
(b)

19

  Matematik Tambahan  Tingkatan 5  Bab 1 Sukatan Membulat 4. Rajah menunjukkan satu sektor SRQ berpusat R

3. Rajah menunjukkan sebuah logo syarikat. ABC dan TS ialah tangen pada S.
ialah segi tiga sama sisi dalam satu bulatan yang
2019 The diagram shows a sector SRQ with centre R and TS
2018 dilukis dengan keadaan AB, BC dan AC adalah is a tangent at S.
tangennya. Terdapat tiga tembereng, P, Q dan R
BAB 1 serupa dilukis dengan keadaan A, B dan C masing- T Q
masing adalah pusat sektornya. P

The diagram shows a logo of a company. ABC is an 11 cm
equilateral triangle in a circle that is drawn such that AB,
BC and AC as tangents. The three similar segments, P, Q 50Њ
and R are drawn such that A, B and C are the centres SR
respectively.

A

P Q Jika panjang lengkok PQ ialah 5 cm, cari
B O
RC If the arc length PQ is 5 cm, find

Jika jejari bulatan ialah 3 cm, cari (a) sudut PRQ dalam radian,

If the radius of the circle is 3 cm, find the angle PRQ in radian,

(a) sudut ABC dalam radian, (b) luas rantau berlorek itu.

the angle ABC in radians, the area of the shaded region.

(b) panjang lengkok AC, (a) 5 = 11q

the arc length AC, q = 5 rad
11
(c) luas rantau berlorek itu.
= 0.45 rad
the area of the shaded region.
T
π P 5 cm Q
3
(a) 60° = rad

A ␪ 11 cm

50Њ
SR

P Q (b) Luas sektor = 1 (11)2(0.45)
O 30Њ 2
Area of sector
B 60Њ C
= 27.2 cm2
(b) AC = 2 3  tan 50° =
tan 30° TS
11

TS = 13.11 cm

= 10.39 cm Luas/Area ∆TSR = 1 (11)(13.11)
2
(c) Luas satu tembereng

Area of a segment = 72.1 cm2
50° 2p = 0.87 rad
= 1 (10.39)2 π  − 1 (10.39)2 sin60° = 50° ×
2 3 2 360°

= 9.78 cm2 Luas sektor SRP = 1 (11)2(0.87)
2
Luas bulatan = π(3)2 Area of sector SRP

Area of circle = 52.6 cm2

Jumlah luas = 9π + 3(9.78) Luas berlorek = 72.1 − 52.6 + 27.2

Total area = 57.61 cm2 Shaded area = 46.7cm2

Praktis
SPM
Ekstra

20

  Matematik Tambahan  Tingkatan 5  Bab 1 Sukatan Membulat 

Sudut KBAT KBAT

Ekstra

Rajah menunjukkan sebuah bulatan berpusat O dan (a) Cari sudut SOP dalam radian. BAB 1
berjejari 10 cm. PQ dan SR adalah dua perentas yang Find the angle of SOP in radian.
selari dengan PQ = 10 cm dan ∠SOR = 120°.
(b) Tentukan luas rantau berlorek itu.
The diagram shows a circle with centre O and radius 10 cm.
PQ and SR are two parallel chords such that PQ = 10 cm and Determine the area of the shaded region.
∠SOR = 120°.
( c) Tunjukkan bahawa perimeter rantau
PQ berlorek ialah
SR 101 + 3 + π  cm.
3
O Show that the perimeter of the shaded region is

  10 1 + 3 + π cm.
3

(a) PQ2 = 102 + 102 − Luas tembereng SR / Area of segment SR
2(10)2 kos POQ
= 1 (10)2 2π  − 1 (10)2 sin 120°
2 3 2
102 + 102 − 102
kos POQ = 2(100) = 61.42 cm2

∠POQ = 60° Maka, luas berlorek

120°−60° Hence, the shaded area
2
∠SOP = = 61.42 – 9.06

π = 52.36 cm2
2
= 30° = rad (c) Panjang lengkok SP = QR

Arc length SP = QR

10 cm 10 π  = 5π
PQ 6 3
SR
SR2 = 102 + 102 − 2(10)2 kos 120°
120Њ
O 10 cm SR = 10 3

Perimeter = 10 + 2 5π  + 10 3
3

(b) Luas tembereng PQ /Area of segment PQ = 101+ π + 3
3

= 1 (10)2 π  − 1 (10)2 sin 60°
2 3 2

= 9.06 cm2

Quiz 1

21

BAB 2 BAB  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan

2 Pembezaan
Differentiation

2.1 Had dan Hubungannya dengan Pembezaan

Limit and its Relation to Differentiation

NOTA IMBASAN

1. Apabila x dalam ungkapan f(x) = 1 menghampiri (c) Jika f(a) = ∞∞ , maka setiap sebutan dalam f(x) perlu
x+2 dibahagikan dengan kuasa tertinggi x. Contoh:

2, nilai x 1 menghampiri 1 , iaitu had x 1 2 = 1 . 1 + x2
in +2 4 x→2 + 4 x2 x2
When x
the expression f(x) = x 1 2 approaches 2, the value of had x2 – 16 = had 1 + x2 = had 2 x2 =
+ x → ∞ x–4 2 – x2 x→∞ x2 x2
x → ∞ – by
f(x) needs =
x +1 2 approaches 1 , that is lim x 1 2 = 1 . 0+1 = −1 example:
4 + 4 0–1
x→2

2. Apabila x menghampiri a, dengan keadaan x ≠ a, maka If f(a) = ∞ , then each term in to be divided
∞
h x →adNa bOagTiAf(xI)MiaBlaAhSL AdaNn ditulis sebagai had f(x) = L. x2 – 16
the highest power of x. For lim x–4
When x approaches to a,where x ≠ a,then the limit for f(x) is L and
x→∞
1 x2
written as lim f(x) = L. + x2
x→a lim 1 + x2 x2 – x2 0+1
x → ∞ 2 – x2 ∞2 x2 0–1
3. Langkah-langkah menentukan had f(x), dengan = lim = = −1
x→
keadaan x→ a
a R adalah seperti x2
∈ berikut:
dy dy δy
Steps to determine lim f(x), where a ∈ R are as follows: 4. Jika y = f(x), dx ditakrifkan sebagai dx = had δx ,
x→a dengan tokokan kecil dalam dx → 0

(a) Jika f(a) ≠ 0 , maka nilai had f(x) = f(a). Contoh: δx ialah x dan δy adalah
h ad (x + 2) 0
= 1 + 2 = 3. x→a tokokan kecil dalam y yang sepadan.
If y = f(x),
x → 1 0 c hange in dy is defined as dy = lim δy , where δx is the small
If f(a) ≠ 0 dx dx dx → 0 δx
, then the value of lim f(x) = f(a). For example:
x→a x and δy is the small change in the corresponding y.


lim (x + 2) = 1 + 2 = 3. dy dy
x → 1 dx dx
(b) Jika f(a) = 0 , maka f(x) perlu diringkaskan dengan 5. Juga, = had .
0 δx → 0


pemfaktoran, atau merasionalkan pengangka atau Also, dy = lim dy .
dx δx → 0 dx

penyebut fungsi itu. Contoh: had x2 – 16 = 6. dy disebut sebagai terbitan pertama y terhadap x.
x–4 dx
x→ 4
had (x – 4)(x + 4) = had(x dy is called the first derivative of y with respect to x.
x→4 + 4) = 8 by dx
x → 4 (x – 4)
f(x) needs to be simplified
If f(a) = 0 , then factorisation, 7. dy juga boleh ditulis sebagai f’(x) apabila y = f(x).
0 dx

or rationalising the numerator or the denominator of the dy
dx
function. For example: can be written as f’(x) when y = f(x).

lim x2 – 16 = lim (x – 4)(x + 4) = lim (x + 4) = 8
x → 4 x – 4 (x – 4)
x→ 4 x→4

22

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan 

1. Cari had bagi setiap fungsi berikut apabila x → 0.

Find the limit for each of the following functions when x → 0. TP 1

CONTOH (a) k x (b) 4 – 2x + x2 BAB 2
2k +
x2 +3x −5
had = k had = 4 – 2x + x2
Penyelesaian: 2k + x x → 0
had x2 + 3x – 5 x → 0 lim
lim = 4
x → 0 x→0 x→0
1 = 2
lim x2 + 3x – 5 = 2

x→0

= 02 + 3(0) – 5
= −5

2. Cari nilai bagi setiap yang berikut.

Find the value for each of the following. TP 2

CONTOH

(i) had (2 – x) (ii) had x2 – 2x – 3 (iii) had (0.02)x
x–3
x → 1 x → 3 x → ∞

lim (2 – x) lim x2 – 2x – 3 lim (0.02)x
x–3
x→1 x→3 x→∞

Penyelesaian:

(i) had (2 – x) = 2 – 1 (ii) h x →ad3 x2 –x 2– x3– 3 = had (x – 3)(x + 1)
x–3
l xim→ 1 x→3
x→1 =1 xli→m 3 = hlxi→mad3 (x + 1) = 4

(iii) had (0.02)x = 0 x→3

x → ∞ lim
x→3
lim
x→∞

(a) had (6 – 2x)  (b) had 1 x (c) had 2
x → 0 3 x → 0 –x+3
lim(6 – 2x) x → ∞ lim x2
1x 2
x→0   lim 3 x→0 x2 –x+3

=6 x→∞

=0 = 2
3

(d) had 2x2 – 8 ( e) x=l h xi→m→ahx2d2→ad22xx222–(–xx62xx–++–331x8 – 4) (f) hxl = xi→m→ahxxl∞di∞→→ma∞d∞14 +1x+4x2x+x+
x → 2 x–2
lim
x→2 2x
2(x2 – 4) lim x
= had x–2 x→2 x
2(x – 4)(x + 1) x
x → 2 2(x + 2)(x – 2) = had x+1
x–2 xxli→→m22 = 0 + 2 = 2
lxi→m2 0 + 1
= had
xli→m2
x→2
=8 = –4

23

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan

3. Cari dy dengan menggunakan prinsip pertama bagi setiap fungsi y = f(x) yang berikut.
dx

Find dy by using the first principle for each of the following functions y = f(x). TP 2
dx

BAB 2 CONTOH

(i) y = 2x + 3 (ii) y = 2
x

Penyelesaian:

(i) y + δy = 2(x + δx) + 3 (ii) y + δy = x 2
+ δx
2x + 3 + δy = 2x + 2δx + 3
(y + δy)(x + δx) = 2
δy = 2δx
yx + yδx + xδy + δx·δy = 2
δy xδy + δx·δy = –yδx
δx = 2 (x + δx)δy = –yδx

M ddyxak = a/hδHxa→edn0c δδexy / dd yx = δy δy = –2
δx δx x(x + δx)
lim
Mddxyak = a/hδHxa→edn0c eδδxy /
δx → 0

= had 2 dy = lim δy
dx δx
δx → 0 δx → 0

= 2 = had = –2 = –2 = –2
δx → 0 x(x + δx) x(x) x2

(a) y = f(x) = 4 − x2 (b) y = f(x) = (x – 2)(x + 1)

δy = f(x + δx) = 4 − x2 x 2 – x – 2
δy = (x + δx)2 – (x + δx) – 2 – x2 + x + 2
= 4 – (x + δx)2 – 4 + x2 = x2 + 2x·δx + (δx)2 – x − δx – x2 + x
= 2xδx – δx + (δx)2
= 4 – x2 – 2xδx – (δx)2 – 4 + x2 δy = [(2x – 1) + δx]δx

= −(2x + δx)δx
δy
δx = –(2x + δx)

had δy = dy = –2x δy = (2x – 1) + δx
δx → 0 δx dx δx
δy dy
had δx = dx = 2x – 1

δx → 0

(c) y = f(x) = 1 (d) y = f(x) = x 1 1
x2 +

δy = 1 = 1 δy = 1 – (x 1 1)
(x + δx)2 x2 (x + δx + 1) +

= x2 – (x + δx)2 = (x + 1) – x – 1 – δx
x2(x + δx)2 (x + 1)(x + 1 + δx)

= –2xδx – (δx)2 = –δx + δx)
δδyx = x2(x + δx)2 δδyx = (x + 1)(x + 1 + δx)
–2x – δx
–1
x2(x + δx)2 (x + 1)(x + 1

had δy = dy = –2x = –2 had δy = –1 = – (x 1
δx dx x4 x3 δx (x + 1)(x + 1) + 1)2
δx → 0 δx → 0

24

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan 

2.2 Pembezaan Peringkat Pertama

The First Derivative

NOTA IMBASAN BAB 2

1. JIkaNyO=TkA, dIeMngBaAn SkeAaNdaan k ialah pemalar, maka dy = 0.
dx
dy
If y = k, where k is a constant, then dx = 0.

2. Jika y = axn, dengan keadaan a dan n adalah pemalar, maka dy = anxn − 1.
dx
dy
If y = axn, where a and n are constants, then dx = anxn − 1.

3. Jika f(x) and g(x) adalah fungsi x dan y = f(x) ± g(x), maka dy = d [f(x) ± g(x)] = df(x) ± dgd(xx). Fungsi ini dibezakan secara
sebutan demi sebutan. dx dx dx

If f(x) and g(x) are functions of x and y = f(x) ± g(x), then dy = d [f(x) ± g(x)] = df(x) ± dg(x) . The function is differentiated term by term.
dx dx dx dx

4. Jika y = g(u) dan u = h(x), maka pembezaan y terhadap x diberi oleh f'(x) = g' (u) × h'(x), iaitu dy = dy × du .
dx du dx

If y = g(u) and u = h(x), then differentiate y with respect to x is given by f‘(x) = g'(u) × h'(x) , that is dy = dy × du .
dx du dx

5. Jika u dan v adalah fungsi x dan y = uv, maka dy = d (uv) = u dv + v du .
dx dx dx dx
dy d dv du
If u and v are functions of x and y = uv, then dx = dx (uv) = u dx +v dx .

  6. u dy dy u v du –u dv
v dx dx v dx dx
Jika u dan v adalah fungsi x dan y = , maka = = .
v2

u dy dy u v du – u dv
v dx dx v dx dx
 
If u and v are functions in x and y = , then = = .

v2

4. Bezakan yang berikut terhadap x.

Differentiate the following with respect to x. TP 2

CONTOH

(i) 5 (ii) 4x2 (iii) 3 Tukarkan kepada
x3 bentuk axn.
Penyelesaian: Change to the form axn.
(i) d(5) = 0
( ii) d(d4xx2) = 4(2)x2 – 1 (iii) 3 = 3x – 3
dx = 8x x3

d(3dxx –3) = 3(–3)x–3 – 1
= –9x – 4

25

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan

(a) 1 (b) x (c) 8x4
2 5

BAB 2 d  1  = 0 d ( x) = d x 1  d  8x4 
dx 2 dx dx 2 dx 5

= 1 = 32 x 3
2x 5

(d) –6 –5x (f) 2x
(2x)3 (e) 3x 4 x

d  –6  = d  –6  d  –5x  = d  –5 x –3   d 2x = d ( – 1
dx (2x)3 dx 8x3 dx 3x4 dx 3 dx x dx
2x 2)

= d  –3 x–3 = 5 =– 2 – 3
dx 4 x4 2 2
x

= 9 = – 2
4x4 2x x

5. Cari f’(x) bagi setiap fungsi yang berikut.

Find f' (x) for each of the following functions. TP 2

CONTOH (a) f(x) = 3
x
f(x) = 1 x5
2
3 – 1
f(x) =
Penyelesaian: x = 3x 2

f’(x) = 1 (5)x 4 Tip – 32 – 3
2
f ’(x) = x 2
5 df(x)
= 2 x4 dx = f'(x) = –3
2x x

(b) f(x) = 10 (c) f(x) = –5
3x –3 4x –2

f(x) = 10 = 10 x3 f(x)= –5 = –5 x3
3x –3 3 4x –2 4

f ’(x) = 10x2 f ’(x) = –5 x
2

26

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan 

6. Cari nilai dy bagi setiap fungsi yang berikut.
dx
dy
Find the value of dx for each of the following functions. TP 2

CONTOH (a) y = 3 x3 , x = 4 BAB 2

y= x5 , x = 2 3
2x2
y = 3x 2

Penyelesaian: ddyx = 9 1
2
x 2

y= x5 = 1 x3 Apabila/When x = 2
2x2 2
dy = 3 (2)2 = 6 = 92 x
dy 1 dx 2
dx = 2 (3)x 2
Tip
dy = 9
dx 2

= 3 x2 Apabila x = 4, (2)
2
Ungkapkan kepada bentuk y = axn. When
Express to the form of y = axn. =9

(b) y = 7x , x = 9 (c) y = –8 , x = – 81
2x 5x –2

7x 7 1 –8 = – 58 x2
2x 2 5x –2
y = = x 2 y =

– 1
ddyx 7 ddxy = – 156 x
= 4 x 2

= 4 7 7 AW pheanbila x = – 81 , ddyx == – 156 – 81 
x 49 2
dy 5
dx


Apabila x = 9, =

When

= 7
12

7. Diberi fungsi polinomial yang berikut, cari terbitan pertama terhadap x bagi setiap yang berikut.

Given the following polynomial functions, find the first derivative with respect to x of each of the following. TP 3

CONTOH

(i) y = 2x3 – 5x2 + 6 (ii) y = (3x – 7)(7 – x)
x
Penyelesaian:
(ii) = (3x – 7)(7 – x)
(i) dy = dy (2x3) – dy (5x2) + dy (6) y x Ringkaskan supaya
dx dx dx dx menjadi sebutan
= 21x – 3x2 – 49 + 7x berasingan dan dalam
= 6x2 – 10x + 0 Beza setiap sebutan x bentuk axn.
= 6x2 – 10x secara berasingan. Simplify to become
Differentiate for each = 28x – 3x2 – 49 separate terms and in
term separately. x the form of axn.

= 28 – 3x – 49x−1

dy = –3 + 49x–2
dx

= –3 + 49
x2

27

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan

(a) y = x5 – 1 x3 + 1 (b) y = – 2 – 1 +2
2 x x5 x2

BAB 2 y = x5 – 1 x3 + x –1 y = –2x – 5 – x – 2 + 2
2
ddyx = 10 + 2
ddxy = 5x4 – 3 1 x6 x3
2 x2 – x2

(c) y = x22 – 3  (d) y = x ( x + 4x2)
x

5

y = 2x2 + 3x y = x + 4x 2

dy = 4x – 3 dy 3
dx dx
= 1 + 10x 2

= 1 + 10x x

(e) y = (2x + 3)(1 – x3) (f) y = 2(x + 1)2
3x 2 3x

y = 2x – 2x4 + 3 – 3x3 y = 2(x2 + 2x + 1)
3x2 3x

= 2 x –1 – 2 x2 + x –2 – x = 2 (x + 2 + x –1)
3 3 3

dy = –2 – 4 x – 1 –1 dy = 2 1 – 1 
dx 3x2 3 x3 dx 3 x2

(g) y = x3(3 – 1 x)2 4x22 – 3 
2 x
(h) y = 7x

y = x39 – 3x + 1 x2 y = 4 x2 – 3 
4 7 x

= 9x3 – 3x4 + 1 x 5 8 12
4 7 7
= x –
dy 5
dx = 27x2 – 12x3 + 4 x4 dy = 8
dx 7
= x227 – 12x + 5 x2
4

28

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan 

8. Cari nilai terbitan pertama untuk fungsi berikut bagi nilai x yang diberikan.

Find the value of the first derivative for each of the following functions for the given value of x. TP 3

CONTOH BAB 2

y =  2 – x 2, x = –1
3

Penyelesaian:

y =  2 – x 2 = 4 – 4x + x2 Apabila/When x = −1,
3 9
dy = 2 (–1 – 2)
dy = – 4 + 2 x dx 9
dx 9 9 = – 2
= 2 (x – 2)
3
9

(a) y = 8 – x2 + x,x=4 5x – 1
3x x
(b) y = , x = –1
1 (2x )2
8
y = 3 x–1 – x2 + x2 5x – x–1 5 1
4x2 4 4
dy –8 1 y = = x–1 – x –3
dx 3x2 2x
= – 2x + dy 5 3
dx 4 4
= – x–2 + x–4

Apabila/When x = 4,

dy = –8 –8+ 1 Apabila/When x = –1,
dx 3(4)2 4
dy = – 5 + 3
= –71112 dx 4 4

= – 1
2

9. Tentukan terbitan pertama bagi fungsi gubahan yang berikut.

Determine the first derivative for the following composite functions. TP 3

CONTOH

(i) (3 – 4x)4 (ii) y = (2x –3 x2)3
Penyelesaian:

(i) Katakan/Let y = (3 − 4x)4 Maka/Hence y = 3u–3

dan/and u = 3 – 4x dy = –9u–4 dan/and
du
Maka/Hence y = u4 dan du = −4
dx du = 2 – 2x
dy = 4u3 dx Kaedah Alternatif

dx dy = dy × du (i) yd=y (3 − 4x)4 − 1· d (3 − 4x)
dx du dx dx = 4(3 − 4x)4 dx
dy = dy × du
dx du dx = –9u–4(2 – 2x) = 4(3 − 4x)3(−4)
= −16(3 – 4x)3
= 4u3(−4)
9 (ii) y = 3(2x – x2)−3
= −16u3 = – u4 (2 – 2x)
dy d
= −16(3 – 4x)3 dx = 3(−3)(2x −x2)−3 − 1· dx (2x − x2)

(ii) Katakan/Let y = 3 = –18(1 – x) = −9(2x – x2)−4(2 – 2x)
– x2)3 (2x – x2)4
(2x = –18(1 – x)
(2x – x2)4
dan/and u = 2x – x2

29

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan

(a) (1 – 4x2)2 (b) 2 – 1 5
3 x

BAB 2 y = (1 – 4x2)2 y = 2 – 1 5
3 x

= 1 (1 – 4x 2)2   dy = 5 2 – 1 4· d (2 – x –1)
3 dx x dx

dy = 2 (1 – 4x2)· d (1 – 4x2) = 52 – 1 4·  1 
dx 3 dx x x2

= 2 (1 – 4x2)(–8x) = 5 2 – 1 4
3 x2 x

= – 16 x(1 – 4x2)
3

(c) 1 (3x3 – 2 x 2)4 (d) 2
2 5
 5+x3

x2

  1 2 4     2 =2 5 –3
y = 2 3x3 – 5 x2 y = x2 +x
5 3
x2 +x
    dy 1 2 d 2
dx = 2 (4) 3x3 – 5 x2 3· dx 3x3 – 5 x 2 5 10
x2 x3
= 23x3 –  2 4 x dy = 2(–3) + x–4– + 1
5 dx
5 x 2 3· 9x2 –

–61 – 10  6(10 – x3)
x3
= =
 5 + x4  x35 +x4
x2 x2

10. Cari nilai dy bagi setiap nilai x yang diberi berikut.
dx

Find the value of dy for each of the following given value of x. TP 4
dx

CONTOH (a) y = 5 2 + x2 , x = 4
4
y= 2 ,x=3
3 4–x
1
Penyelesaian: 1 y = 5 (2 +
2 4 x2) 2
–
y= 2 = 2 (4 – Apabila/When x = 3,   dy = 5 1 – 1
4–x 3 x) dx 4 2
3 dy 1 (2 + x2) 2 (2x)
dx = 3(1)
Jika/If u = 4 – x, du = –1, 1 = 5x
dx 4 2 + x2
1 1
y= 2 – 2 = 3
3
u dy 5(4)
dx 4 18
dy 1 – 3 Apabila x = 4, =
du 3 2 When 5
=– u
52
dy = dy × du = 32 = 6
dx du dx
Tip
=– 1 ·(–1)
31
3
u 2 = u·u 2
3u 2
= 1

3(4 – x) 4 – x

30

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan 

(b) y = (x4 − 3x2 − 2)3, x = −1 (c) y = (x 6 x 2) , x = –2
+

y = (x4 – 3x2 – 2)3 y = 6(x + x2)–1

dy = 3(x4 – 3x2 – 2)2(4x3 – 6x) dy = –6(x + x 2)–2(1 + 2x) BAB 2
dx dx

= 6(2x3 – 3x)(x4 – 3x2 – 2)2 = –6(1 + 2x)
(x + x2)2
Apabila x = –1, dy = 6(–2 + 3)(1 – 3 – 2)2
When dx = 96 dy –6(–3)
Apabila x = –2, dx = (–2 + 4)2

When = 9
2

11. Tentukan terbitan pertama bagi suatu fungsi yang melibatkan hasil darab ungkapan algebra yang berikut.

Determine the first derivative for the following functions which involve the multiplication of algebraic expressions of the

following. TP 4

CONTOH (a) y = x3(1 − 2x)2

y = (x – 2)2(3 – x2)3 y = x3(1 – 2x)2

u = x3 v = (1 – 2x)2

Penyelesaian: du = 3x2 dv = 2(1 – 2x)(–2)
dx dx
y = (x – 2)2(3 – x2)3

Kata/Let u = (x – 2)2 v = (3 – x2)3 dd yx == x3(2)(–2)(1 – 2x) + (1 – 2x)2(3x2)
x2(1 – 2x)[–4x + 3(1 – 2x)]
du = 2(x – 2) dv = 3(–2x)(3 – x2)2
dx dx
= x2(1 – 2x)(3 – 10x)
= –6x(3 – x2)2
dy dv vddux
dx = u dx +

= (x – 2)2(–6x)(3 – x2)2 + (3 – x2)3(2)(x – 2)

Ambil faktor yang sama.
Take the same factor.

= 2(x – 2)(3 – x2)2[–3x(x – 2) + 3 – x2]
= 2(x – 2)(3 – x2)2[6x + 3 – 4x2]

(b) y = (x + 4)2(1 −3x)2 (c) y = (2x − 1)(1 + x2)4

y = (x + 4)2 (1 – 3x)2 y = (2x –1)(1 + x2)4

dy = (x + 4)2(2)(–3)(1 – 3x) + (1 – 3x)2(2)(x + 4) dy = (2x – 1)(4)(2x)(1 + x2)3 + (2)(1 + x2)4
dx dx
= 2(1 – 3x)(x + 4)[–3(x + 4) + 1 – 3x] = 2(1 + x2)3[4(2x – 1)(x) + 1 + x2]

= 2(1 – 3x)(x + 4)[–6x – 11] = 2(1 – x2)3[9x2 – 4x + 1]

31

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan

(d) y = x2 x – 1 (e) y = 2(x − 2)(1 − 3x)3

1 y = 2(x – 2)(1–3x)3

y = x2(x – 1) 2 dy = 2(x – 2)(3)(–3)(1 – 3x)2 + 2(1 – 3x)3
BAB 2 – 1 dx
  dy = x2 1 + x – 1(2x) = 2(1 – 3x)2[–9(x – 2) + 1 – 3x]
dx 2 (x – 1) 2

= x2 + 2x x – 1 = 2(1 – 3x)2[19 – 12x]
2 x–1

= x2 + 4x(x – 1)
2 x–1

= 5x2 – 4x
2 x–1

12. Tentukan terbitan pertama bagi suatu fungsi yang melibatkan hasil bahagi ungkapan algebra yang berikut.

Determine the first derivative for the following functions which involve the division of algebraic expressions. TP 4

CONTOH (a) y = (x – 3)3
(1 + 2x)2

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

Penyelesaian: dy = (1 + 2x)2(3)(x – 3)2 – (x – 3)3(2)(2)(1 + 2x)
y = (1 + 2x)2 dx (1 + 2x)4

(1 – x)3 (1 + 2x)(x – 3)2[3(1 + 2x) – 4(x – 3)
(1 + 2x)4 3
Katakan u = (1 + 2x)2, v = (1 – x)3 =

Let du = 4(1 + 2x) dv = –3(1 – x)2 = (x – 3)2[15 + 2x)
dx dx (1 + 2x)3

dy v du –u dv
dx dx v2 dx
Gunakan =

Use

dy = (1 – x)3(4)(1 + 2x) – (1 + 2x)2 (–3)(1 – x)2
dx (1 – x)6

= (1 – x)2(1 + 2x)[4(1 – x) + 3(1 + 2x)]
(1 – x)6

= (1 + 2x)[4 – 4x + 3 + 6x]
(1 – x)4

= (1 + 2x)[7 + 2x]
(1 – x)4

32

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan 

(b) y = (1 x2 (c) y = 1 + 2x2
– 2x)3 2 + x3

y = (1 x2 y = 1 + 2x2 BAB 2
– 2x)3 2 + x3

dy = (1 – 2x)3(2x) – x2 (3)(–2)(1 – 2x)2 dy = (2 + x 3)(4x) – (1 + 2x2)(3x2)
dx (1 – 2x)6 dx (2 + x3)2

= 2x(1 – 2x)2[1 – 2x + 3x] = x[4(2 + x3) – 3(1 + 2x2)x]
(1 – 2x)6 4 (2 + x3)2

= 2x(1 + x) = x(8 + 4x3 – 3x – 6x3
(1 – 2x)4 (2 + x3)2

= x[8 – 3x – 2x3]
(2 + x3)2

(d) y = 2x3 – 3 (e) y = x x
(x – 1) +1

2x3 – 3 1
(x – 1)
y = y = x = x2
x+1 x+1
(x – 1)(6x2) – (2x3 – 3)
dy = (x – 1)2 (x + 1) – x
dx 2x
dy (x + 1)2
= 4x3 – 6x2 + 3 dx =
(x – 1)2
= x + 1 – 2x
NOTA IMBASAN 2 x (x + 1)2

= 1 – x
2 x (x + 1)2

2.3 Pembezaan Peringkat Kedua

The Second Derivative

NOTA IMBASAN

1. Jika y = f(x), dy = f'(x) ialah pembezaan peringkat pertama.
dx

If y = f(x), dy = f'(x) is the first order differentiation.
dx

  d dy = d2y = f''(x) ialah pembezaan peringkat kedua
dx dx dx2

  d dy = d2y = f''(x) is the second order differentiation.
dx dx dx2

  2. d2y dy 2.
Perhatikan bahawa dx2 ≠ dx

  d2y dy 2.
Note that dx2 ≠ dx

33

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan

13. Cari terbitan pertama dan kedua bagi fungsi berikut yang berkenaan dengan x. Seterusnya, bandingkan bentuk
 2 dd2xy2 dandy 2
dx

BAB 2 Find the first and the second derivatives for the following functions with respect to x. Then, compare the expressions of
 d2y dy
and dx 2. TP 3
dx2

CONTOH (a) y = (x − 3)2(4 − x2)

y = 4x3 – 2x2 + 1 x ddyx = (x – 3)2(–2x) + (4 – x2)(2)(x – 3)
2 = 2(x – 3)[–x(x – 3) + 4 – x2]
= 2(x – 3)(3x + 4 – 2x2]
Penyelesaian:

dy = 12x2 – 4x + 1 dd2xy2 = (2x – 6)(–4x + 3) + (– 2x2 + 3x + 4)(2)
dx 2 = –8x2 + 6x + 24x – 18 – 4x2 + 6x + 8
= –12x2 + 36x – 10
d2y = 24x – 4
dx2

 dy 2 = 12x2 – 4x + 1 2   ddyx 2 = 4(x – 3)2(3x + 4 – 2x2]2
dx 2

Maka/Hence ∴ dd2xy2 ≠ dy
  dx 2
d2y ≠  dy 2
dx2 dx

(b) y = 1 – 1 + 2x2 (c) y = x
x x2 2x + 1

y = x–1 – x–2 + 2x2 ddyx = (2x + 1) – 2x = 1
(2x + 1)2 (2x + 1)2
ddxy = – x–2 + 2x –3 + 4x
dd2xy2 = –4
dd2xy2 = 2x–3 – 6x–4 + 4 (2x + 1)3

= 2 – 6 +4  ddyx 2 = 1
x3 x4 (2x + 1)4

∴ dd2xy2 ≠ dy
 ddyx 2 = 4x – 1 2 2   dx 2
x2
+ x3

∴ dd2xy2 ≠ dy
  dx 2

34

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan 

14. Jawab semua soalan yang berikut.

Answer all the following questions. TP 4

CONTOH (a) Diberi y = x2 + 2x. Tunjukkan bahawa BAB 2

  2x2 d2y − dy 2 + 4 dy = 4.
dx2 dx dx
Diberi y = 2x – 1 . Tunjukkan bahawa x2 d2y +
x dx2   d2y dy 2+
x2 + 2x. Show that 2x2 dx2 − dx


4x dy  = 12x − 2y Given y =
dx
1 d2y dy 4 dy = 4.
 Given y = 2x – x dx2 + 4x dx dx
. Show that x2

= 12x − 2y. y = x2 + 2x

Penyelesaian: y’ = 2x + 2

y = 2x – x –1 y’’ = 2

dy = 2 + 1 Sebelah kiri = 2x2[2] – (2x + 2)2 + 4(2x + 2)
dx x2
Left side = 4x2 – [4x2 + 8x + 4] + 8x + 8

d2y = –2 = 4 sebelah kanan
dx2 x3
right side

   Sebelah kiri = x2 d2y + 4x dy
= x2 dx dx
Left side 2 1
x3 x2

– + 4x 2 +

=– 2 + 8x + 4
x x

= 2 + 8x
x

Sebelah kanan = 12x – 2y
 Right side
= 12x – 2 2x – 1
12x – 4x + x
2
x
=

= 8x + 2
x

 ∴ d2y dy = 12x – 2y
x2 dx2 + 4x dx

(b) Diberi y = x3 – 2x2. Tunjukkan bahawa (c) Diberi y = x + 2 . Tunjukkan bahawa d2y +
x dx2
x2 d2y – 6y = 8x2. 2 dy
dx2 x dx + x = y.

Given y = x3 – 2x2. Show that x2 d2y – 6y = 8x2. Given y = x + 2 . Show that d2y + 2 dy + x = y.
dx2 x dx2 x dx

y = x3 – 2x2 y = x + 2x –1
y' = 1 – 2x –2
y’ = 3x2 – 4x y'' = 4x–3

y’’ = 6x – 4

Sebelah kiri = x2[6x – 4] – 6[x3 – 2x2]   Sebelah kiri = 4 2 2
x3 x x2
Left side = –4x2 + 12x2 + 1– +x

= 8x2 sebelah kanan Left side 2
x
right side = +x

sebelah kanan/right side

35

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan

2.4 Aplikasi Pembezaan

Application of Differentiation

BAB 2 NOTA IMBASAN

1. Tafsiran dy sebagai kecerunan lengkung y = f(x).
dx

Defining dy as the gradient of a curve y = f(x).
dx

(a) Katakan P(x, y) dan titik terdekat Q(x + δx, y + δy) adalah dua titik pada lengkung y = f(x), maka kecerunan PQ = δy .
δx
δy
Let P(x, y) and the nearest point Q(x + δx, y + δy) are two points on the curve y = f(x), then the gradient PQ = δx .

y

Q (x+δx, y+δy) Tangen pada P
Tangent at P

P (x, y) ␦y
␦x
x
0

(b) Jika kecerunan tangen pada P ialah m, maka m = δy , iaitu had δy = dy =m
δx δx → 0 δx dx

If the gradient of the tangent at P is m, then m = δy , that is lim δy = dy =m
δx δx → 0 δx dx

2. Kecerunan ItoMafnthBgeeAtnaSn, mgAe1Nn×t, kecerunan normal, m2 = −1. m2 = −1.
m1 × the gradient of the normal,
TheNgrOadTieAnt

3. Persamaan tangen atau normal boleh ditentukan dengan menggunakan rumus (y – y1) = m(x − x1), dengan m ialah
kecerunan tangen atau normal.

The equation of the tangent or normal can be obtained by using the formula (y – y1) = m(x − x1), where m is the gradient of the tangent or the
normal.

4. Satu titik pada lengkung y = f(x) yang mempunyai dy = 0 dinamakan titik pusingan. Pada titik ini, tangen adalah selari
dengan paksi-x. dx

A point on the curve y = f(x) whose dy = 0 is known as the turning point. At this point, the tangent is parallel to the x-axis.
dx

5. y δy y

δy δx = 0 δy
δx > 0 δx < 0
P δy δy
δx < 0 δy δx > 0

δx = 0

0 x < x1 x1 x > x1 x 0 x < x1 Q x>x x
(i) 1
x1
(ii)

Pada rajah (i), titik P merupakan titik maksimum. Pada rajah (ii), titik Q merupakan titik minimum. Perhatikan tanda dy di
sebelah kiri dan sebelah kanan masing-masing bagi titik P dan Q. Note the signs of on the left and right of the turning dx
dy
In diagram (i), P is the maximum point. In diagram (ii), Q is the minimum point. dx points
P and Q.

6. Langkah berikut boleh digunakan untuk menentukan titik maksimum atau minimum bagi suatu lengkung y = f(x).

The following steps can be used to determine the maximum or minimum point for a curve y = f(x).

36

NOTA IMBASAN   Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan 

(i) Cari/Find dy
dx

(ii) Cari nilai x apabila dy = 0. BAB 2
dx

Find the value of x when dy = 0.
dx

(iii) Tentukan sama ada titik pusingan (x, y) adalah titik maksimum atau minimum dengan menggunakan
Determine whether the turning point (x, y) is a maximum point or minimum point by using the

(a) jadual/table

Nilai x Pilih satu nilai x lebih Nilai x pada titik Pilih satu nilai x lebih Titik pusingan
Value of x kecil daripada nilai x pusingan. besar daripada nilai x Turning point
pada titik pusingan. The value of x at the pada titik pusingan.
Choose a value of x smaller turning point. Choose a value of x larger
than the value of x at the than the value of x at the
turning point. turning point.

Nilai dy Jika/If dy  0, x  x1 dy =0 Jika/If dy  0, x  x1 Minimum
dx dx dx dx Maksimum/maximum

Value of dy Titik refleks
dx Reflex point

NOTA IMBASAN

(b) terbitan kedua

second derivatives (x1, y1), jika d2y  0 pada x = x1, maka titik tersebut ialah titik maksimum.
Pada titik pusingan dx2 the x = x1, then the point is a maximum point.
d2y
, y1), if dx2  0 at

NAOt TthAe tuIMrniBngApSoiAntN(x1

Pada titik pusingan (x1, y1), jika d2y  0 pada x = x1, maka titik tersebut ialah titik minimum
dx2
d2y
At the turning point (x1 , y1), if dx2  0 at the x = x1, then the point is a minimum point

Pada titik pusingan (x1, y1), jika d2y = 0 pada x = x1, maka titik tersebut ialah titik reflek.
dx2
d2y
At the turning point (x1 , y1), if dx2 = 0 at the x = x1, then the point is a reflex point.

7. Kadar perubahan yang terhubung

Rate of related change dy yang diberi oleh petua rantai dy = dy × dx .
Jika y = f(x) dan x = g(t), maka kadar perubahan y terhadap masa, t ialah dt dt dx dt

If y = f(x) and x = g(t), then the rate of change of y with respect to time, t is dy which is obtained by the chain rule dy = dy × dx .
dt dt dx dt

8. Tokokan kecil dan penghampiran

Small changes and approximation

Kita diketahui bahawa had δy = dy .
δx → 0 δx dx

We know that lim δy = dy .
dx → 0 δx dx


Jika δy dan δx adalah sangat kecil, maka δy adalah satu hampiran bagi dy , iaitu δy ≈ dy → δy ≈ dy × δx.
δx dx δx dx dx

If δy and δx are small changes, then δy is an approximation for dy , that is δy ≈ dy → δy ≈ dy × δx.
δx dx δx dx dx

37

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan

BAB 2 15. Cari fungsi kecerunan bagi setiap lengkung yang diberi dan seterusnya cari nilai kecerunan pada titik x yang

diberi dan nilai koordinat y yang sepadannya.

Find the gradient function for the following curves and then find the value of the gradient at the given value of point x with

its corresponding coordinates of y. TP 3

CONTOH (a) y = x(5 − 3x) pada/at x = −2.

y = x3 − 2x pada x = 2. y = 5(–2) – 3(–2)2
x = –10 – 12 = –22
y = x3 − 2x at x = 2. y' = 5 – 6x

Penyelesaian:

Fungsi kecerunan, dy = 3x2 − 2 Pad a/At x = –2 ddyx = 5 – 6(–2)
dx = 17

The gradient function, dy = 3x2 − 2
dx

Apabila/When x = 2, y = 23 − 2(2) = 4

dy = 3(2)2 – 2 = 10
dx

Maka, kecerunan lengkung pada titik (2, 4) ialah
10.

Hence, the gradient of the curve at point (2, 4) is 10.

(b) f(x) = 4x2 – 6x + 1 pada/at x = –1. (c) f(x) = −2x + 4x2 pada/at x = 1.

f(–1) = 4(–1)2 – 6(–1) + 1 f(1) = –2 + 4 = 2
= 11 f'(x) = –2 + 8x
f'(x) = 8x – 6 f'(1) = –2 + 8(1) = 6
f'(–1) = 8(–1) – 6 = –14 ∴ Pada (1, 2), f'(x) = 6
∴ Pada (–1, 11), f'(x) = –14 At

At

38

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan 

16. Cari nilai p dan q bagi yang berikut.

Find the value of p and of q for the following. 4

CONTOH (a) Kecerunan lengkung y = p + qx pada (2, 4) BAB 2
x
Kecerunan lengkung y = px + qx2 pada (1, 1) ialah ialah −8.
−5. p
The gradient for the curve y = x + qx at (2, 4)
The gradient for the curve y = px + qx2 at (1, 1) is −5. is −8.

Penyelesaian: y = px–1 + qx

Pada/At (1, 1), 1 = p(1) + q(1)2 ... ➀ p
2
dy = p + 2qx 4 = + 2q
dx
y' = –p +q
Maka/Hence p + 2q(1) = −5 ... ➁ x2

➁ − ➀, q = −6 Apabila x = 2 –4p + q = –8

Gantikan dalam ➀, When

Substitute into ➀, –p + 4q = –32 ...➁

p=1+6=7 ➁ + ➂ p + 4q = 8 ...➂

8q = –24
q = –3
p = (4 + 6)2
= 20

(b) Kecerunan lengkung y = px3 + qx pada (−1, 1) (c) Kecerunan lengkung y = px2 + qx + 1 pada
ialah 5. (2, 5) ialah 8.

The gradient for the curve y = px3 + qx at (−1, 1) is 5. The gradient for the curve y = px2 + qx + 1 at (2, 5)
is 8.

y = px3 + qx (–1, 1) y = px2 + qx + 1 (2, 5)
1 = –p – q ...➀ 5 = 4p + 2q + 1
y' = 3px2 + q 2 = 2p + q ...➀
y' = 2px + q,
Apabila x = –1 y' = 3p + q = 5 ...➁

When ➁ + ➀ 2p = 6 Apabila x = 2 y' = 4p + q = 8 ...➁

p = 3, q = –4 When ➁ – ➀ 2p = 6

p = 3, q = –4

39

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan

17. Cari persamaan tangen dan normal kepada lengkung pada titik dengan koordinat-x yang diberi.

Find the equation of the tangent and the normal to the curve at the point with the given coordinates of x. TP 4

BAB 2 CONTOH (a) y = 1 pada/at x = 2
x

y = 3 – 2x – x2 pada/at x = −2. 1 1
x 2
Penyelesaian: y = x = 2, y =

y = 3 – 2(−2) – (−2)2 = 3 dy = – 1
dx x2
dy = −2 − 2x
dx dy 1
Pada x = 2, dx =– 4
Maka, kecerunan tangen pada x = −2 ialah
Persamaan tangen/Equation of tangent
Hence, the gradient of tangent at x = −2 is

−2 – 2(−2) = 2 1 1
2 4
Persamaan tangen pada (−2, 3) y– =– (x – 2)

The equation of the tangent at (−2, 3) 1
4
y – 3 = 2[x – (−2)] y = – x + 1

y – 3 = 2x + 4 Persamaan normal//Equation of normal

y = 2x + 7 1
2
Kecerunan normal/Gradient of the normal y – = 4(x – 2)

=– 1 m1 × m2 = –1 y = 4x – 15
2 2

Persamaan normal Kaedah Alternatif

pada (−2, 3). Persamaan tangen y = 2x + c pada

Equation of the normal titik (–2, 3)
at (−2, 3). 3 = 2(–2) + c
c=7
1 Persamaan normal y = – 1 x + c
y – 3 = – 2 (x + 2) 2
pada titik (–2, 3)

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

y = – 1 x + 2 c1 = +2
2
∴y = – 1 x + 2
2

(b) y = x3 – 4x pada/at x = –1 (c) y = x pada/at x = 4

y = x3 – 4x x = –1 1

y = –1 – 4(–1) y = x = x 2 x = 4, y = 2

= 3 dy 1
dx 2x
dy = 3x2 – 4 pada x = –1 =
dx
dy 1
ddxy == 3(–1)2 – 4 Pada x = 4, dx = 4
–1
Persamaan tangen/Equation of tangent
Persamaan tangen/Equation of tangent
y – 2 = 1 (x – 4)
y – 3 = –(x + 1) 4

y = –x + 2 y = 1 x + 1
4
Persamaan normal/Equation of normal

y – 3 = x + 1 Persamaan normal//Equation of normal

y = x + 4 y – 2 = –4(x – 4)

y = –4x + 18

40

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan 

18. Selesaikan masalah yang melibatkan tangen dan normal. BAB 2

Solve the problems involving tangents and normals. TP 4

CONTOH

Diberi f(x) = x − 1 . Cari koordinat bagi titik jika f'(x) = 2. Seterusnya, cari persamaan tangen pada titik
dengan x  0. x

Given that f(x) = x − 1 . Find the coordinates of the point if f'(x) = 2. Then, find the equation of the tangent at the point
whose x  0. x

Penyelesaian:

f(x) = x – 1 Persamaan tangen pada (1, 0)
x Equation of the tangent at (1, 0)

f'(x) = 1 + 1 =2 y – 0 = 2(x – 1)
x2
1 y = 2x – 2
x2
= 1 Persamaan normal pada (1, 0)
Equation of the normal at (1, 0)
x = ±1
y – 0 = – 1 (x – 1)
Apabila /When x = 1 f(1) = 1 – 1 = 0 2

x = –1 f(–1) = –1 + 1 = 0 y = – 1 x + 1
2 2
(1, 0) dan (–1, 0)

(a) Kecerunan titik P yang terletak pada lengkung y = 3x2 − 6x adalah selari dengan garis lurus y = 12x − 3.

The gradient of point P on the curve y = 3x2 − 6x is parallel to the line y = 12x − 3.

Cari/Find

(i) koordinat P,

the coordinates of P,

(ii) persamaan normal pada P.

the equation of the normal at P.

(i) y = 3x2 – 6x (ii) Persamaan normal pada (3, 9)
Equation of the normal at (3, 9)
dy
dx = 6x – 6 = 12 y – 9 = – 1 (x – 3)
12
6x = 18 x = 3
y = 3(3)2 – 6(3) y = – 1 x + 37
12 4
= 9

∴P[3, 9]

(b) Cari koordinat bagi titik-titik pada lengkung y = 2x + 8 , dengan tangennya selari dengan paksi-x.
x
8
Find the coordinates of the points on the curve y = 2x + x , such that the tangents are parallel to the x-axis.

y = 2x + 8
x

dy = 2 – 8 =0
dx x2

2x2 = 8 x = ±2

y = 2(2) + 8 , 2(–2) – 8
2 2

= 8, –8

∴(2, 8), (–2, –8)

41

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan

(c) Tangen kepada lengkung y = ax2 + bx + 2 pada titik 1, 1  adalah selari dengan normal kepada
2

lengkung y = x2 + 6x + 4 pada (−2, −4). Cari nilai a dan nilai b.

BAB 2   1, 1 is parallel to the normal for the curve y = x2 + 6x + 4 at
The tangent to the curve y = ax2 + bx + 2 at the point 2

the point (−2, −4). Find the value of a and of b.

y = ax2 + bx + 2 ∴2a + b12 = =–a21+ .b..➀+

dy = 2ax + b 2
dx

pada (1, 1 ), dy = 2a + b a + b = – 3 ...➁
2 dx 2

y = x2 + 6x + 4 ➀ – ➁ a = 1, b = –1

ddxy = 2x + 6 = – 5
p ada (–2, –4), ddyx == 2
2(–2) + 6
2

19. Cari titik pusingan bagi lengkung dan tentukan sifat-sifat bagi titik tersebut.

Find the turning point(s) of the curves and determine the characteristics of these points. TP 4

CONTOH

y = 6x – x2 Kecerunan dy menukar tanda dari positif ke
dx
Penyelesaian:
negatif semasa melalui x = 3, jadi titik pusingan itu
dy
dx = 6 − 2x ialah maksimum.

Untuk titik pusingan, kecerunan dy =0 The gradient dy changes from positive to negative
dx dx
as it passes through x = 3, hence the turning point is a
dy
For the turning point, gradient dx =0 maximum point.

Maka/Then 6 − 2x = 0

x = 3 Kaedah Alternatif

y = 18 − 9 = 9 d2y
Gunakan terbitan kedua, jadi dx2 = −2
Titik pusingan ialah (3, 9).
The turning point is (3, 9).
Use second derivative, so d2y = −2
Dari jadual berikut. dx2
d2y
From the following table. Apabila dx2  0, titik pusingan (3, 9) ialah maksimum.

Sedikit Sedikit When d2y < 0, the turning point (3, 9) is maximum
x kurang 3 3 lebih 3 dx2

Less than 3 More than 3

dy + 0−
dx

Lakaran

tangen

Sketch of the
tangent

42

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan 

(a) y = 2x2 − 4x − 1

dy = 4x − 4 x Sedikit 1 Sedikit
dx kurang 1 0 lebih 1
dy dy BAB 2
Untuk titik pusingan, kecerunan dx =0 dx − +

For the turning point, gradient dy =0 Lakaran
dx tangen

Maka, 4x − 4 = 0

Then

x = 1 dan y = 2 − 4 − 1 dy
dx
= −3 Kecerunan menukar tanda dari negaif ke

Titik pusingan ialah (1, –3) positif semasa melalui x = 1, jadi titik pusingan itu

The turning point is (1, –3) ialah minimum.

Dari jadual berikut. The gradient dy changes from negative to positive
dx
From the following table.

as it passes through x = 1, hence the turning point is

minimum.

(b) y = x3 − 3x2 (c) y = 1 x 3 − x2 − 3x + 1
3

dy = 3x2 − 6x dy = x2 − 2x – 3
dx dx
dy dy
Untuk titik pusingan, kecerunan dx =0 Untuk titik pusingan, kecerunan dx =0

For the turning point, gradient dy =0 Maka x2 − 2x – 3 = 0
dx
(x – 3)(x + 1) = 0
Maka 3x2 − 6x = 0
x = 3 dan –1
x = 0 dan 2
y= –8 dan 8
y = 0 dan −4 3
8
Titik pusingan ialah (0, 0) dan (2, −4) Titik pusingan ialah (3, –8) dan −1, 3 

The turning point is (0, 0) and (2, −4) Dari jadual berikut.

Dari jadual berikut./From the following table.

x Sedikit 0 Sedikit x Sedikit 0 Sedikit
kurang 0 0 lebih 0 kurang 3 lebih 3
dy dy
dx − 2 + dx − 0+
Lakaran 0 Lakaran
tangen Sedikit Sedikit tangen
kurang 2 lebih 2
x x Sedikit –1 Sedikit
+ – kurang –1 lebih –1
dy
dx dy + 0–
Lakaran dx
tangen
Lakaran
tangen

(0, 0) ialah titik minimum dan (2, −4) ialah (3, −8) ialah titik minimum dan −1, 8  ialah
3
titik maksimum. titik maksimum.
8 ) is a
(0, 0) is a minimum point and (2, –4) is a maximum (3, –8) is a minimum point and (–1, 3
point. maximum point.

43

BAB 2   Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan

20. Selesaikan masalah yang melibatkan nilai maksimum dan nilai minimum yang berikut.

Solve the following problems involving the maximum and minimum values. TP 5

CONTOH

Sebuah tangki segi empat yang tertutup = 27 + 4x2
hendak dibuat supaya dapat mengisi 9 m3 air. x
Panjang tangki mesti dua kali lebarnya. Cari
dimensi tangki itu jika jumlah luas permukaan ddAx = − 27 + 8x =0
ialah minimum. x2

A closed rectangular tank is made to fill 9 m3 of Untuk minimum/For minimum
water. The length of the tank is twice its breadth.
Find the dimensions of the tank if the total surface 8x = 27
area is minimum. x2

Penyelesaian: x3 = 27
8

x = 3
2

hm d2A = 54 +8
dx2 x3

xm Apabila/When

2x m x= 3 , d2A  0 (minimum)
2 dx2
Isi padu/Volume = 9 m3 = 2x2h
9
Jumlah luas permukaan/Total surface area h= =2
2 9 
A = 2(2xh) + 2xh + 4x2 4

= 6xh + 4x2 Panjang/Length = 3 m, Lebar/Breadth = 3 m,
9 2
= 6x 2x2  + 4x2 Tinggi/Height = 2 m

(a) Rajah menunjukkan satu kad jemputan. Ia terdiri daripada satu segi empat tepat dengan panjang y cm
dan lebar x cm. Pada kedua-dua hujung segi empat tepat itu, terdapat satu segi tiga sama sisi dengan
panjang sisi x cm.

The diagram shows an invitation card. It is made up of a rectangle with a length of y cm and width of x cm. At the
both ends of the rectangle, there are equilateral triangles with sides of x cm.

y cm

x cm 60Њ

Jaipkaabpilearximbeetrenrilkaei s4el–u13ru3hacnmr.ajah ialah 26 cm, buktikan bahawa luasnya adalah satu nilai maksimum

If the perimeter of the whole diagram is 26 cm, prove that the area is maximum when the x value is 13 cm.
4– 3

Perimeter = 26 = 2y + 4x Untuk A maksimum, dA = 0
dx
y = 13 – 2x For a maximum A

 Luas A = xy + 1 60° 2 dA = 13 – 4x + 2x sin60° = 0
2 dx
Area A
x2sin

= x[13 – 2x] + x2sin60° 13 = x4 – 23 
2

∴x = 13
4– 3

44

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan 

(b) Seutas dawai 360 cm panjang digunakan untuk membuat 12 sisi sebuah kotak yang berbentuk segi BAB 2

empat tepat dengan panjangnya adalah dua kali lebarnya. Jika lebar kotak ialah x cm,

A wire of length 360 cm is used to make 12 sides of a rectangular box whose length is twice its breadth. If the
breadth is x cm.

(i) buktikan bahawa isi padu kotak, V diberi oleh V = 180x2 – 6x3.

prove that the volume of the box, V is given by V = 180x2 – 6x3.

(ii) seterusnya tentukan dimensi kotak supaya isi padunya bernilai maksimum.

then determine the dimensions of the box so that its volume is maximum.

(i) 4t + 8x + 4x = 360 t
t + 3x = 90
I = 2x2t x
= 2x2[90 – 3x] 2x
= 180x2 – 6x3

(ii) ddxI = 360x – 18x2 = 0
x[360 – 18x] = 0

x = 0 x = 20

Dimensi 20 × 40 × 30 cm

Dimension

(c) Rajah menunjukkan sekeping kadbod ABCD, dengan BC = 6 cm, AB = 10 cm dan AD = 26 cm. Ahmad
ingin menggunting sebuah segi empat AEFG dari kadbod supaya luasnya adalah maksimum.

The diagram shows a piece of cardboard ABCD, where BC = 6 cm, AB = 10 cm and AD = 26 cm. Ahmad wants to
cut out a rectangle AEFG from the cardboard so that the area of the rectangle is maximum.

BC

GF
x cm

Jika FE ialah x cm, A ED

If FE is x cm,

(i) nyatakan ED, AE dan luas AEFG dalam sebutan x.
state ED, AE and the area of AEFG in terms of x.

(ii) Seterusnya, cari luas maksimum segi empat itu.

Then, find the maximum area of the rectangle.

(i) 1x0 = ED (ii) dA = 26 – 4x = 0
20 dx

ED = 2x x= 13 cm
2

∴AE = 26 – 2x    Luas maksimum = 13 26 – 2 13
= 2 2
Luas AEFG = x[26 – 2x] Maximum area
84 1
Area AEFG 2 cm2

45

BAB 2   Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan

21. Bagi setiap persamaan yang berhubung dengan x dan y berikut, jika kadar perubahan x ialah 0.5 unit s−1, cari
kadar perubahan y pada ketika yang diberi. Tafsirkan jawapan anda.

For each of the following equations relating x and y, if the rate of change of x is 0.5 unit s−1, find the rate of change of y at the

given instant. Interpret your answers. TP 3

CONTOH

y = 2x2 + 1, x = 1.2 Maka/Hence dy = 4(1.2)(0.5)
dt
Penyelesaian:
= 2.4 unit s−1
dx
Diberi/Given dt =0.5units−1apabila/when x = 1.2 Kadar perubahan y ialah 2.4 unit s−1.
The rate of change of y is 2.4 unit s−1.
dy = 4x
dx
dy dy dx
Maka/Hence dt = dx × dt Tip

= 4x dx Oleh sebab dy  0, perubahan y dikatakan
dt bertambah. dt

Apabila/when x = 1.2 dan/and dx = 0.5 Since dy 0, the change of y is increasing.
dt dt


(a) y = (3x − 6)2, x = 1 (b) y = x x 1 , x = 4
–

Diberi dx = 0.5 unit s−1 apabila x = 1. DGiivbeenr i ddxt = 0.5 unit s−1 awphaebnila x = 4.
dt
Given When

dy = 6(3x − 6) dy = (x – 1) – x
dx dx (x – 1)2

Maka dy = dy × dx = –1
dt dx dt (x – 1)2
Hence
= 6(3x − 6) dx   Maka dy = dy × dx
dt dt = dx dt
Hence
Apabila x = 1 dan dx = 0.5 –1 dx
dt dt
When and (x – 1)2

Maka dy = 6(−3)(0.5) Apabila x = 4 dan dx = 0.5
dt dt
When and
Hence = −9 unit s−1
dy –1
Kadar perubahan y ialah −9 unit s−1. Maka dt = (4 – 1)2 (0.5)

y menyusut. Hence 1
The rate of change of y is –9 unit s–1. 18
= – unit s−1
y decreases.
1
Kadar penyusutan y ialah 18 unit s−1

The rate of y decreasing is 1 unit s–1
18

46

  Matematik Tambahan  Tingkatan 5  Bab 2 Pembezaan  BAB 2

22. Bagi setiap persamaan yang berhubung x dan y berikut, jika kadar perubahan y ialah 8 unit s–1, cari kadar
perubahan x pada ketika yang diberi. Tafsirkan jawapan anda.

For each of the following equations relating x and y, if the rate of change of y is 8 unit s–1, find the rate of change of x at the

instant given. Interpret your answers. TP 3

CONTOH

y = x2 − 6x, x = −1 Apabila/When x = −1 dan/and dy =8
dt
Penyelesaian:
6] dx 
Diberi/Given dy = 8 unit s−1 apabila/when x = −1 Maka/Hence 8 = [2(–1) – dt
dt
dy dx 8
dx = 2x – 6 dt = –8 unit s−1

dy dy dx = –1 unit s−1
dt dx dt
Maka/Hence = × Kadar perubahan x ialah –1 unit s−1
The rate of change of x is –1 unit s−1
= (2x – 6) ×  dx 
dt

Tip

Oleh sebab dx < 0, perubahan x dikatakan
menyusut. dt

Since dx  0, the change of x is decreasing.
dt

(a) y = 4x2 , x = 3 (b) y = 2x – 3, x = 2
x–1

Diberi dx = 8 unit s−1 apabila x = 3 DGiivbeenr i ddyt = 8 unit s−1 awphaebnila x = 2
dt
Given when

dy = (x – 1)(8x) – 4x2 dy = 1
dx (x – 1)2 dx 2x – 3

dy = dy dx dy dy dx
 Maka dt dx × dt Maka dt = dx × dt

Hence Hence
= (x – 1)(8x) – 4x2 dx 1  dx 
dt = 2x – dt
(x – 1)2 3

Apabila x = 3 dan dy =8 Apabila x = 2 dan dx =8
dt dt
When and When and
dx
(3 – 1)(8)(3) – 4(3)2 dx 1 dt 
(3 – 1)2 dt
 Maka 8 = Maka 8 =

Hence Hence

dx = 8 unit s−1 dx = 8 unit s−1
dt 3 dt

Kadar perubahan x ialah 8 unit s−1. Kadar perubahan x ialah 8 unit s−1.
3
x bertambah.
x bertambah. 8
3 The rate of change of x is 8 unit s–1.
The rate of change of x is unit s–1. x increases.

x increases.

47