Hybrid PBD 2022 Tg 5 - Matematik Tambahan

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan
18. Cari nilai h dan nilai k bagi yang berikut. SP 2.4.1 TP4
Find the value of h and of k for the following.
h
Kecerunan lengkung y = hx – 2kx pada titik (3, 1) (a) Kecerunan lengkung y = + kx pada (1, –1)
2
x
ialah –4. ialah -5.
h
The gradient for the curve y = hx – 2kx at the point (3, 1) is –4. The gradient for the curve y = + kx at (1, –1) is –5.
2
x
Pada (3,1) , 1 = h(3) – 2k(3)
2
1 = 3h – 18k …… Pada/At (1, –1) , –1 = h + k(1)
d y = h – 4kx –1 = h + k ……. 
dx d y = –h + k
dx x 2
Maka/hence h – 4k(3) = –4
h = 12k – 4 …… Maka/Hence –h + k = – 5 ………
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Gantikan  dalam , 1 = 3(12k – 4) – 18k  + , 2k = –6
Substitute  into , 18k = 13 k = –3
13 h = –1+3
k = , = 2
18
13
h = 12   – 4 = 14
3
18

(b) Kecerunan lengkung y = hx + kx pada (2, 0) (c) Kecerunan lengkung y = hx + kx – 3 pada (–2, 9)
2
3
ialah –8. ialah –7.
The gradient for the curve y = hx + kx at (2, 0) is –8. The gradient for the curve y = hx + kx – 3 at (–2 ,9) is –7.
3
2
Pada/At (2, 0) , 0 = h(8) + k(2) Pada/At (–2, 9), 9 = 4h + k(–2) – 3
0 = 4h + k ……. 6 = 2h – k ……. 
d y = 3hx + k d y = 2hx + k
2
dx dx
maka/hence 3h(2) + k = –8 maka/hence 2h(–2) + k = –7
2
12h + k = –8 ……. -4h + k = –7 ……. 
 – , 8h = –8  + , –2h = –1
h = –1 h =
1
k = 4 2
k = –5




19. Cari persamaan tangen dan normal kepada lengkung pada titik yang diberi berikut. SP 2.4.2 TP4
Find the equation of tangent and normal to the following curves at the given points.


f(x) = x + 3x – 6 pada titik (–1, 5).
2
3
f(x) = x + 3x – 6 at the point (–1, 5).
3
2
f’(x) = 3x + 6x
Apabila/When x = –1, f’ (–1) = 3(–1) + 6(–1)
2
= –3
1
Kecerunan tangen pada titik (–1, 5) ialah –3. Kecerunan normal pada titik (–1, 5) ialah .
Gradient of the tangent at point (–1, 5) is –3. Gradient of the normal at point (–1, 5) is . 3
1
Persamaan tangen ialah 3
Equation of the tangent is Persamaan normal ialah / Equation of the normal is
1
y – 5 = –3(x – (–1)) y – 5 = (x – (–1))
3
y – 5 = –3x – 3 3y – 15 = x + 1
y = –3x + 2
3y – x = 16





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02 Hybrid PBD Mate Tamb Tg5.indd 30 09/11/2021 9:24 AM

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan

(a) f(x) = 2x – 5x + 7 pada titik (3, 2).
3
2
3
2
f(x) = 2x – 5x + 7 at the point (3, 2).
f’ (x) = 6x – 10x
2
Apabila/When x = 3, f’ (3) = 6(3) – 10(3) Kecerunan normal pada titik (3, 2) ialah – 1 .
2
= 24 Gradient of the normal at point (3, 2) is – 1 24
.
Kecerunan tangen pada titik (3, 2) ialah 24. 24
Gradient of the tangent at point (3, 2) is 24. Persamaan normal ialah / Equation of the normal is
Persamaan tangen ialah y – 2 = – 1 (x – 3)
Equation of the tangent is 24
y – 2 = 24(x – 3) 24y – 48 = –x + 3
y – 2 = 24x – 72 24y + x = 51
y = 24x – 70
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
(b) f(x) = 3x + 1 pada titik (1, 0).

f(x) = 3x + 1 at the point (1, 0).
3
f’(x) =

2 3x + 1
3 3 4
Apabila/When x = 1, f’(1) = = . Kecerunan normal pada titik (1, 0) ialah – .

2 3(1) + 1 4 4 3
3
Kecerunan tangen pada titik (1, 0) ialah . Gradient of the normal at point (1, 0) is – .
3
3 4
Gradient of the tangent at point (1, 0) is . Persamaan normal ialah / Equation of the normal is
4
4
Persamaan tangen ialah y – 0 = – (x – 1)
Equation of the tangent is 3
4
3 y = – x + 4
y – 0 = (x – 1) 3 3
4
3 3
y = x –
4 4
20. Selesaikan masalah berikut yang melibat persamaan tangen dan normal kepada lengkung yang berikut.
Solve the following problems involving the tangent and the normal to the following curves. SP 2.4.2 SP 2.4.3 TP6

Rajah menunjukkan satu jalan raya mengikut persamaan y = x – 4x. Sebuah y
3
kereta bergerak pada jalan itu dengan keadaan bahawa apabila di A(–1, 3),
arah halaju mengikut tangen kepada jalan pada A. Cari persamaan tangen
itu. y = x – 4x
3
The diagram shows a road that follows the equation y = x – 4x. A car moves on the road such that 0 x
3
when it is at A(–1, 3), the direction of the velocity is the tangent to the road at that point A. Find the
equation of the tangent.
Kecerunan lengkung ialah d y = 3x – 4
2
The gradient of curve is dx
Di titik/At point A(–1, 3) , d y = 3(–1) – 4 = –1
2
dx
Maka, persamaan tangen/Hence, the equation of tangent
y – 3 = –1[x – (–1)]
y = –x + 2


Di/At A( –1, 3) , 3 = –1(–1) +c
c = 2
Maka/Hence Gunakan y = mx + c bagi persamaan tangen
Use y = mx + c for the equation of tangent
y = –x + 2







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02 Hybrid PBD Mate Tamb Tg5.indd 31 09/11/2021 9:24 AM

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan

(a) y (b) Rajah menunjukkan satu y
bahagian trek larian
A mengikut persamaan y y = √ x – 1
C P

20 m = x – 1. Tom berada di P
0 P Q 160 m x dan arah halajunya ialah 0 x
tangen pada P, dengan Q
Rajah menunjukkan sebuah jejambat berbentuk persamaan 4y – x = 3.
parabola dan tertinggi di A. Cari The diagram shows a part of the running track whose
The diagram shows a parabolic shaped flyover with the equation is y = 
x – 1. Tom is at P and his velocity direction
maximum height at A. Find is the tangent at P given by 4y – x = 3.
(i) persamaan bagi jejambat. (i) Cari koordinat di P./Find the coordinates of Tom at P.
the equation of the flyover. (ii) Pada ketika Tom di P, kawannya Ramin di Q,
(ii) tangen bagi arah sebuah kereta di C apabila dengan keadaan PQ adalah berserenjang
OP = PQ. dengan arah Tom. Cari persamaan PQ dan
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the tangent of the direction of a car at C when OP= PQ koordinat Q.
When Tom is at P, his friend Ramin is at Q, such that
(i) y = a(x – 80) + 20 PQ is perpendicular to the direction of Tom. Find the
2
Apabila/When x = 0, y = 0 equation of PQ and the coordinates of Q.
1
0 = a(–80) +20 (i) Diberi/Given y = (x – 1) …..
2
2
–1 4y = x + 3
a =
320 d y 1 –1 2
–1 dx = (x – 1)
2
y = (x – 80) +20
2
320 maka/hence
1
–1
d y –2 (x – 1) = 1
2
(ii) = (x – 80) 2 4 1
dx 320 2 = (x – 1) 2
Apabila/When x = 40, d y = –1 (–40) = 1 x = 5, y = 2
dx 160 4
P(5, 2)
Persamaan tangen/Equation of tangent (ii) Persamaan PQ ialah y – 2 = –4(x – 5)
1
y – 15 = (x – 40) y = –4x + 22 …….
4
1 Dari  dan 
y = x + 5 1 2
4 (x – 1) = 22 – 4x
16x – 177x + 485 = 0
2
(16x – 97)(x – 5) = 0
97 –9
x = , 5 y = , 2
16 4
,
Q  97 –9 
16 4
21. Cari titik pusingan bagi lengkung dan tentukan sifatnya bagi yang berikut dengan kaedah (i) memerhatikan
kecerunan titik-titik kejiranannya (ii) terbitan kedua. SP 2.4.4 TP4
Find the turning point(s) of the curves and determine the nature of these points by (i) observing the gradients of the neigbouring points (ii)
second derivatives.
Simulasi
y = 2x – 4x – 1 x – 1 1 2
2
Kaedah I d y –8 0 4
d y = 4x – 4 dx
dx Tanda bagi d y
Untuk mencari titik pusingan, kecerunan d y dx – 0 +
To find turning point, the gradient Sign for dx
d y = 0 Lakaran tangen
dx Sketch of tangent
maka/then 4x – 4 = 0 Dari jadual / From the table
x = 1 dan y = 2 – 4 – 1 = –3
Titik pusingan ialah (1, –3) Kecerunan d y menukar tanda dari negatif ke positif semasa
Turning point is dx
melalui x = 1, jadi titik pusingan itu ialah minimum.
dy
The gradient changes from negative to positive as it passes through x = 1,
dx
hence the turning point is a minimum point
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02 Hybrid PBD Mate Tamb Tg5.indd 32 09/11/2021 9:24 AM

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan

Kaedah II
2
Gunakan terbitan kedua, d y = 4
dx 2
Use second derivative, we get
2
Apabila d y > 0, titik pusingan (1, –3) ialah minimum
dx 2
2
d y
When > 0, the turning point (1, –3) is minimum.
dx 2

1
(a) y = –x + 2x + 6 (b) y = x – 2x – 5x + 1
2
2
3
d y = –2x + 2 d y 3 2
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dx dx = x – 4x – 5
(i) Untuk titik pusingan, kecerunan d y = 0 (i) Untuk titik pusingan, kecerunan d y = 0
For the turning point, gradient dx For turning point, gradient d y = 0 dx
Maka/Hence –2x + 2 = 0 2 dx
Maka/Hence x – 4x – 5 = 0
x = 1 dan y = 7 (x – 5)(x + 1) = 0
Titik pusingan ialah (1, 7) 1 11
Turnig point is x = 5 dan –1 y = –32 dan 3
3
x –1 1 2 Titik pusingan ialah 5, –32 1 dan –1, 11
d y  3   3 
dx 4 0 –2 Dari jadual / From the table
d y x 4 5 6
Tanda bagi d y
d y dx + 0 – –5 0 7
Sign for dx
dx d y
Lakaran tangen Tanda bagi dx – 0 +
Sketch of tangent Sign for d y
dx
Dari jadual/From the table Lakaran tangen
Kecerunan d y menukar tanda dari positif ke Sketch of tangent
dx x –2 –1 0
negatif semasa melalui x = 1, jadi titik pusingan d y
itu ialah maksimumn dx 7 0 –5
The gradient dy/dx changes from positive to negative as it Tanda bagi d y
passes through x = 1, hence the turning point is a maximum d y dx + 0 –
point. Sign for dx
2
d y Lakaran tangen
(ii) = –2 Sketch of tangent
dx 2 1 11
2

Oleh sebab d y 2 , 0, titik pusingan (1, 7) ialah  5, –32 3  ialah titik minimum dan –1, 3  ialah
dx
titik maksimum maksimum.

3 
3 
 5, –32 1 is a minimum point and –1, 11 is maximum.
Simulasi
2
(ii) d y = 2x – 4
dx 2 2
Apabila x = 5, d y = 6 (.0)
dx 2
d y 1
2

Oleh sebab dx 2 . 0, titik pusingan 5, –32 3 
ialah titik minimum.
d y
2
Apabila x = –1, = –6(,0)
dx 2
d y 11
2

Oleh sebab dx 2 , 0, titik pusingan –1, 3  ialah
titik maksimum.



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02 Hybrid PBD Mate Tamb Tg5.indd 33 09/11/2021 9:24 AM

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan
22. Selesaikan masalah yang berikut. SP 2.4.5 TP6
Solve the following problems.


Cikgu Wong ingin balik rumah secepat mungkin dari suatu tempat Rumah/ House R
A, 4 km dari stesen bas, S dan rumahnya, R ialah 7 km dari stesen
bas seperti ditunjukkan dalam rajah. Jika dia boleh berjalan
dengan halaju 3 km sejam dari A ke P dan berbasikal dengan 5 km P 7 km
sejam dari P ke R, cari jarak P dari stesen bas, S.
Mr Wong wants to go home as soon as possible from a place A, which is 4 km from the Stesen bas/ Bus station
bus station S and 7 km from the bus station to his house, R as shown in the diagram. A 4 km S
If he can walk with a velocity of 3 km per hour from A to P and cycles at 5 km per hour
from P to R, find the distance of P from the bus station, S.
Tip Penting
Katakan jumlah masa diambil ialah T dan jarak SP ialah x km. Semakan/Check
Let the total time taken be T and the distance SP be x km.  2 7 – 3
2
4 + 3
Maka T = masa diambil dari A ke P berjalan + masa dari P ke R T(3) = 3 + 5 = 2.47
4 + 3
berbasikal T(–3) =  2 + 7 – (–3) = 3.67 jam
2
Then T = time taken from A to P walking + time taken from P to R cycling 3 5
2
4 + x
 2 7 – x
= +
3 5
Untuk menentukan jumlah masa T minimum, d T = 0
For the total minimum time T, dx
Maka/Then
d T 1(2x) 1
= – = 0
dx 3.2 4 + x 5
 2
2
5x = 3 16 + x
 2
25x = 9x + 144
2
2
16x = 144
2
x = 3, atau –3 tidak sesuai (ke arah selatan)
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not suitable (go to south)
Maka, masa paling singkat ialah apabila x = 3.
Hence, the shortest time taken is when x = 3.
(a) Cari tinggi, t dan jejari, j sebuah silinder tertutup yang paling murah dibina dengan j j
mempunyai isi padu 1 000 cm . Harga bahan untuk membina dasar dan tapak ialah
3
2
2
RM0.05 cm dan harga bahan untuk sisi ialah RM0.03 cm .
Find the height, t and radius, j of the cheapest closed cylinder that has a volume of 1 000 cm . The material t t
3
cost to make the top and the base is RM 0.05 cm while the cost for the side is RM0.03 cm .
2
2
Isi padu/Volume, V = pj t = 1 000 ……
2
1 000
t =
pj 2
2
Jumlah kos, C = 0.05pj (2) + 0.03(2pjt)
Total cost = 0.1pj + 0.06pjt …….
2
2 
2
2
= 0.1pj + 0.06pj  1 000 = 0.1pj + 60
j
pj
dC = 0.2pj – 60 = 0

dj j 2
0.2pj = 60
3
1 000
j = 4.57 cm, t =
p(4.57) 2
= 15.24 cm




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02 Hybrid PBD Mate Tamb Tg5.indd 34 09/11/2021 9:24 AM

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan

(b) Anda diminta menentukan perjalanan paling murah Jalan raya/ Road 8 km Bandar/ Town
bagi kabel telefon yang menyambung sebuah sekolah S Kabel x Ladang kelapa sawit 2 km
ke bandar O seperti ditunjukkan dalam rajah. Cable Palm oil estate
You are asked to determine the cheapest route for a telephone cable that Sekolah/ School
connects a school S to the town O as shown in the diagram.
(i) Jika kos menyambung kabel di tepi jalan ialah RM5 000 sekilometer dan RM8 000 sekilometer
melalui ladang kelapa sawit, cari kos terendah.
If the cost to lay the cables by the road side is RM5 000 per km and RM8 000 per km across the palm oil estate, find the least
cost.
(ii) Apakah kos terendah jika harga kos kabel ialah RM7 000 sekilometer ditambah kepada pembinaan
tersebut.
What is the least cost if the cable cost is RM7 000 per km is added to cost to lay the cables.
(i) Kos/Cost C = (8 – x)5 000 + 8 000 2 + x
 2
2
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dC 8 000(2x)
= –5 000 +

dx 2 4 + x 2
dC
Untuk minimum, = 0
dx
4 000(2x)
5 000 = 
4 + x
2
100 = 64x – 25x 2
2
100
x =  = 1.6 km
39
Kos terendah/The least cost C = (8 – 1.6)5 000 + 8 000 2 + 1.6
2
 2
= RM52 490
(ii) Panjang kabel/Length of cable = (8 - 1.6) + 2 + 1.6 = 8.96 km
 2
2
Jumlah kos/Total cost = 8.96 × RM7 000 + RM52 490 = RM115 210
23. Bagi setiap persamaan yang berhubung dengan x dan y berikut, jika kadar perubahan x ialah 0.04 unit s , cari
–1
kadar perubahan y pada ketika nilai x yang diberi. Tafsirkan jawapan anda. SP 2.4.6 TP4
For each of the equation relating x and y, if the rate of change of x is 0.04 unit s , find the rate of change of y at the given value of x. Interpret
–1
your answers.
(a) y = (2x + 3) , x = –1 x
3
1
y = x + 6 , x = (b) y = x – 2 , x = 3
2
2 d x
d x Diberi = 0.04 unit/s apabila d x
Diberi = 0.04 unit/s apabila Given dt when Diberi dt = 0.04 unit/s apabila
Given dt when x = –1 Given when
1 d y x = 3
x = = 3(2x + 3) (2) d y [(x – 2) – x] –2
2
2
d y = 2x dx dx = (x – 2) 2 = (x – 2) 2
dt Maka/Then dy = d y × d x d y d y d x
Maka/Then dt dx dt d x Maka/Then dt = dx × dt
2
d x
d y = × d x = 2x   = 6(2x + 3) dt –2 2  
d x
dt dx dt dt Apabila x = –1 dan d x = 0.04 = (x – 2) dt
1 dt
Apabila/When x = Apabila/When x = 3
2
d x Maka/Then Maka/Then
dan/and = 0.04 d y 2 d y –2
dt dt = 6(2(–1) + 3) (0.04) = (0.04)
Maka/Then = 0.24 unit/s dt (3 – 2) 2
1
d y = 2   (0.04) = 0.04 unit/s y bertambah dengan kadar = –0.08 unit/s
dt 2 0.24 unit/s y menyusut dengan kadar
y bertambah dengan kadar y increases at a rate of 0.24 unit/s y decreases at a rate of
y increases at a rate of 0.08 unit/s
0.04 unit/s




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02 Hybrid PBD Mate Tamb Tg5.indd 35 09/11/2021 9:24 AM

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan
24. Bagi setiap persamaan yang berhubung x dan y berikut, jika kadar perubahan y ialah 0.08 unit s , cari kadar
–1
perubahan x pada ketika nilai x yang diberi. Tafsirkan jawapan anda. SP 2.4.6 TP4
For each of the following equations relating x and y, if the rate of change of y is 0.08 unit s , find the rate of change of x at the instant given
–1
value of x. Interpret your answers.

1 (b) y = 5x + 6, x = 2
x 2 (a) y = x – 4x , x = –1
2
y = , x = 4 2
(x – 1) d y
Diberi / Given Diberi d y = 0.08 unit/s Diberi dt = 0.08 unit/s
dt
d y = 0.08 unit/s apabila/ when apabila x = –1 apabila x = 2
dt d y = 5 
5x + 6
x = 4 d y = x – 4 dx 2
d y = [(x –1)(2x) – x ] dx
2
dx (x –1) 2 d y d y d x Maka/Then d y = d y × d x
dx
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dt
dt
Guna petua rantai/Use chain rule Maka/Then dt = dx × dt
d x
d x
5x + 6
d y = d y × d x = (x – 4)   = 5   
dt
2
dt dx dt dt
2
[(x –1)(2x) – x ] d x Apabila/When x = –1 dan Apabila x = 2 dan d y = 0.08
=   dt
(x –1) 2 dt d y
= 0.08
dt
 
Apabila/When x = 4 dan/and Maka/Then Maka/Then 0.08 = 5 d x
dy 8 dt
d x
dt = 0.08 0.08 = (–1– 4)   d x = 0.128 unit/s
Maka/Then dt dt
[(4 – 1)(2(4) –(4) ] d x d x 0.08
2
0.08 =   = unit/s x bertambah dengan kadar
(4 – 1) 2 dt dt –5 0.128 unit/s.
= –0.016 unit/s
9
dx = 0.08 × unit/s x menyusut dengan kadar x increases at a rate of 0.128 unit/s.

dt 8
= 0.09 unit/s 0.016 unit/s.
x decreases at a rate of 0.016 unit/s.
x bertambah dengan kadar
0.09 unit/s.
x increases at a rate of 0.09 unit/s.

25. Selesaikan masalah yang berikut. SP 2.4.7 TP6
Solve the following problems.
Semasa dipanaskan, panjang jejari, j cm bagi sebiji bola kuprum bertambah dengan kadar malar 2%. Cari
While heating, the radius, j cm of a copper ball increases uniformly at 2%. Find
(i) peratus perubahan dalam isi padu.
the percentage change in the volume.
(ii) peratus perubahan dalam luas permukaan bola kuprum.
the percentage change in the surface area of the copper ball.
Tafsirkan nilai peratus perubahan tersebut.
Interpret the value of change of the percentage.
(i) Diberi/Given δj × 100% = 2 %
j
4
Isi padu bola kuprum V = pj Tip Penting
3
Volume of copper ball 3 • Jejari, j cm bagi sebuah bola kuprum bertambah
dV = 4pj 2 dengan malar 2% ialah δj j × 100% = 2%.
dj
δV dV δj The radius, j cm of a copper ball increases uniformly
Maka/Then × 100% = × × 100% δj
V dj V at 2% is j × 100% = 2%.
δj
= 4pj × × 100% • Peratus perubahan dalam isi padunya ialah
2
4pj 3 Percentage change in the volume is defined by
3 δV × 100%
δj v
= 3 × 100% = 3(2%) = 6%
j
Isi padu bola kuprum bertambah 6% apabila jejarinya bertambah 2%
The volume of the copper ball increases by 6% when the radius increases by 2%.

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02 Hybrid PBD Mate Tamb Tg5.indd 36 09/11/2021 9:24 AM

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan

(ii) Luas permukaan bola kuprum
Surface area of the copper ball Tip Penting
A = 4pj 2 • Peratus perubahan dalam luas permukaannya ialah
dA = 8pj Percentage change in the surface area is defined by
dj
δA dA δj δA × 100%
Maka/Then ×100% = × × 100% A
A dj A
8pjdj

= × 100%
4pj 2
δj
= 2 ×100% = 2(2%) = 4%
j
Luas permukaan bola kuprum bertambah 4% apabila jejarinya bertambah 2%.
The surface area of the copper ball increases by 4% when the radius increases by 2%.
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(a) Sebuah kon tegak dengan bucunya ke bawah, sudut bucunya ialah 60°. Jika kon itu diisi dengan
3
minyak sehingga penuh dan minyak kemudiannya dibiar mengalir keluar dengan kadar malar 3 cm /s,
cari peratus perubahan tinggi minyak di dalam kon, dalam sebutan p ketika ketinggian minyak di
dalam kon ialah 6 cm.
A vertical cone with its tip at the bottom, the angle at the tip is 60°. If the cone is fully filled with oil and then the oil is allowed to drip
out from the tip at a constant rate of 3 cm /s, find the change of the percentage of the height of oil in the cone, in terms of π when
3
the height of the oil inside the cone is 6 cm.
Katakan jejari ialah j dan tinggi ialah h. / Let say the radius is j and the height is h.
tan 30 = j
0
h h
 3 = j
1 1 1 1

V = pj h = p  h h = p h 3 60°
2
2
3 3 3 9
dV = ph 2
1
dh 3
dV × 100% = dV × dh × 100%
dh dh V
1 dh
–3 = ph ×  3 × 100%
2
3 1 ph
9
dh
–3 = 3 × h × 100%
dh
× 100% = –1%
h
(b) Luas sebuah bulatan bertambah dengan kadar malar 4%. Tentukan
The area of a circle increases uniformly at 4%. Determine
(i) peratus perubahan jejarinya. (ii) peratus perubahan lilitannya.
the percentage change in the radius. the percentage change in the circumference.
δA Tafsirkan nilai peratus perubahan tersebut.
(i) Diberi/Given × 100% = 4% Interpret the value of change of the percentage.
A
Luas bulatan/Area of circle
A = pj 2 (ii) Lilitan bulatan/Circumference
dA = 2pj L = 2pj
dj dL = 2p
δA dA δj dj
Maka/Then × 100% = × × 100%
A dj A Maka/Then δL × 100% = dL × δj × 100%
2pj δj L dj L
4% = × 100%
pj 2 2p dj
δj 4% = 2pj × 100%
× 100% = = 2%
i 2 = 2 %
Jejari bulatan bertambah sebanyak 2% apabila Lilitan bulatan bertambah sebanyak 2%
luasnya bertambah sebanyak 4%. apabila jejarinya bertambah sebanyak 2%.
The radius increases by 2% when the area increases by 4 %. The circumference increases by 2% when the radius
increases by 2%



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02 Hybrid PBD Mate Tamb Tg5.indd 37 09/11/2021 9:24 AM

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan
26. Tentukan nilai hampir bagi setiap yang berikut. SP 2.4.8 TP5
Determine the small change for each of the following.

2 2
Diberi y = x – 2x, cari tokokan kecil dalam y apabila (a) Diberi y = 2 , cari nilai hampir bagi 2 .
2
x berubah daripada 2 kepada 2.01. Seterusnya cari (2x – 1) 2 (3.02)
nilai hampir bagi y. Given that y = (2x – 1) 2 , find the approximate value
2
2
Given y = x – 2x, find the small change in y when x changes from 2 of (3.02) 2 .
to 2.01. Then, find the approximate value of y.
Dengan bandingan: 2x – 1 = 3.02
Diberi/Given y = x – 2x By comparison 2x = 4.02
2
dy = 2x – 2 x = 2.01
dx Apabila x berubah daripada 2 kepada 2.01,
x berubah daripada 2 kepada 2.01 When x changes from 2 to 2.01, δx = 0.01
x changes from 2 to 2.01. 2
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Jadi/Then δx = 2.01 – 2 = 0.01 Dari/From y = (2x – 1) 2
δy dy dy –8
≈ Tip Penting =
δx dx δy dy dx (2x – 1) 3
dy ≈ dy
δy ≈ × δx δx dx Maka/Hence δy ≈ × δx
dx δy ≈ dy × δx dx
≈ (2x – 2) δx dx ≈ –8 δx
≈ (2(2) – 2) (0.01) (2x – 1) 3
≈ 0.02 Apabila x = 2 dan δx = 0.01
Apabila/When x = 2, y = 2 – 2(2) δy ≈ –8 (0.01) ≈ -0.00296
2
= 0 (2(2) – 1) 3 2 2
Maka, nilai hampir bagi y = y + δy Apabila/When x = 2 , y = (2(2) – 1) 2 = 9
= 0 + (0.02) 2
= 0.02 Maka, nilai hampir bagi (3.02) 2
Hence, the approximate value of y Hence, the approximate value of 2
2 (3.02) 2
= + (–0.00296) = 0.219
9
3 dy dy
(b) Diberi y = , cari . Seterusnya, cari nilai (c) Diberi y = (5x – 2) , cari dan nilai hampir bagi
3
(x + 1) 2 dx dx
1 3.05 .
3
hampir bagi .
(4.01) 2 Given y = (5x – 2) , find dy and the approximate value of 3.05 .
3
3
3
Given y = (x + 1) 2 , find dy . Then, find the approximate value dx
dx
1 Diberi/Given y = (5x – 2) 3
for .
(4.01) 2 dy = 15(5x – 2) 2
Diberi/Given y = 3 dx
(x + 1) 2 5x – 2 = 3.05
dy –6
= 5x = 5.05
dx (x + 1) 3 Apabila x berubah daripada 1 ke 1.01.
Jika/If x + 1 = 4.01 When x changes from 1 to 1.01
x = 3.01 Jadi/Then δx = 1.01 – 1 = 0.01
x berubah daripada 3 kepada 3.01 δy dy
x changes from 3 to 3.01 δx ≈ dx
Jadi/Then δx = 3.01 – 3 = 0.01 dy
δy ≈ dy δy ≈ dx × δx
δx dx 2
dy –6 ≈ 15(5x – 2) δx
δy ≈ × δx ≈ δx
dx (x + 1) 3 Apabila/When
–6
≈ (0.01) ≈ –0.00094 x = 1
(3 + 1) 3 δy ≈ 15(5(1) – 2) (0.01)
2
3
Apabila/When x = 3 , y = 16 ≈ 1.35
Maka, nilai baharu/ Hence, the new value Nilai hampir bagi 3.05 = (5 – 2) + 1.35 = 28.35
3
3
y = y + δy = 3 + (-0.00094) = 0.1866 Approximate value for
16
1
Nilai hampir bagi 1 2 = ( 0.1865) = 0.0622
Approximate value for (4.01) 3

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02 Hybrid PBD Mate Tamb Tg5.indd 38 09/11/2021 9:24 AM

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan
27. Selesaikan yang berikut. SP 2.4.9 TP5
Solve the following.



Mangkuk terbuka berbentuk hemisfera dengan Luas permukaan minyak,
jejari 13 cm. Minyak mengalir ke dalamnya dan Area of surface of oil
tinggi minyak dalam mangkuk berubah daripada A = pR 2
2
2
2
3 cm kepada 3.05 cm. Cari perubahan hampir luas Tetapi/But R = 13 – (13 – h) = 26h – h 2
2
permukaan minyak dalam mangkuk itu, dalam Maka/Hence A = p(26h – h )
sebutan p. dA = p(26 – 2h)
An open hemispherical bowl has a radius of 13 cm. Oil is flowing dh
into the bowl such that the height of the oil in the bowl increases δA ≈ dA .δh
from 3 cm to 3.05 cm. Find the approximate change of the surface dh
area of the oil in the bowl in terms of p. ≈ p(26 – 2h) δh
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Apabila/When δh = 0.05 dan h = 3
δA ≈ p(26 – 2(3)) (0.05)
≈ p
2
h R Jadi, luas permukaan bertambah sebanyak p cm .
Then, the surface area increases by p cm . 2


(a) Isi padu, V cm bagi suatu kon menyusut daripada 30 cm kepada 29.5 cm . Jika tinggi kon sentiasa
3.
3
3
tiga kali jejari tapak kon itu, cari
The volume, V cm of a cone decreases from 30 cm to 29.5 cm . If the height of the cone is always three times the length of the
3
3
3
base radius of the cone, find
(i) perubahan hampir tinggi apabila jejari tapak ialah 1.2 cm.
the approximate change in the height when the base radius is 1.2 cm.
(ii) perubahan hampir jejari jika tinggi kon ialah 6 cm.
the approximate change in the base radius when the height is 6 cm.
1 1 1
(i) h = 3j V = pj h = pj 3 (ii) V = pj h = ph 3
2
2
3 3 27
δV = 29.5 – 30 = –0.5 cm 3 dV 1
dV = 3 pj 2 dh = ph 2
9
dj dV
dV δV = dh .δh
δV = .δj
dj 1 2
–0.5 –0.5 = p(6) δh
2
–0.5 = 3 pj .δj δj = = –0.0368 9
3p(1.2) 2 –0.5
δh =
4p
Dari/From h = 3j
dh
dj = 3
dh
δh = dj .δj
–0.5 = 3 δj
4p
δj = –0.013



















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02 Hybrid PBD Mate Tamb Tg5.indd 39 09/11/2021 9:24 AM

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan

(b) Salah satu sisi sebuah kuboid telah diukur oleh seorang murid. Dia tidak sengaja tertulis panjang sisi
sebagai 4.4.cm walaupun panjang ialah 4.5 cm. Disebabkan oleh kesalahan ini, hitung
One of the length of a cuboid was measured by a student. Instead of recording the length as 4.5 cm, he accidently wrote 4.4 cm.
Because of this error, calculate
(i) perubahan hampir bagi isi padu kuboid.
the approximate change in the volume of the cuboid.
(ii) perubahan peratus bagi luas permukaan kuboid.
the percentage change in the surface area of the cuboid.

(i) Diberi/Given δx = 4.4 - 4.5 = –0.1 (ii) Luas/Area A = 6x 2
V = x dA
3
dV = 3x 2 dx = 12x
dx dV δA = dA × δx
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δV = × δx dx
dx
= 3(4.5) (–0.1) = 12(4.5)(–0.1) = –5.4
2
2
= –6.075 Luas sebenar = 6(4.5) = 121.5
Peratus perubahan = –5.4 ×100%
Percentage change 121.5
= –4.44%



28. Lakukan projek STEM di bawah. SP 2.4.5 TP6 Nota Visual
Cary out the STEM project below.

Project-Based Learning
ojek
Pr
Projek
Objektif aktiviti: Menyelesaikan masalah yang melibatkan maksimum atau minimum dengan
Activity objective: pembezaan.
Solve the problems involving maximum or minimum by differentiation.
Pernyataan masalah: Apakah jarak terpendek dari suatu titik kepada satu garis lurus atau suatu lengkung
Problem statement: yang diberi?
What is the shortest distance from a given point to a given line or curve?

Pencarian fakta: Gunakan pembezaan dan ciri-ciri titik pusingan untuk mencari jarak minimum dan
Fact finding: jarak maksimum.
Use differentiation and the characteristics of turning point to find the minimum or maximum distance.
Konsep yang Syarat untuk menentukan nilai maksimum atau minimum bagi kalkulus, iaitu
diaplikasikan: dy = 0 bagi dua pemboleh ubah, x dan y.
Concept applied: dx
The condition to determine the maximum or minimum value in calculus, that is dy = 0 for two variables,
x and y. dx
Bahan yang Persamaan garis lurus dan lengkung
diperlukan: Kertas graf
The equation of the straight line and curve
Graph papers
Pelan Tindakan/ Action plan:
1. Bahagikan kelas kepada beberapa kumpulan terdiri daripada 4 orang.
Divide the class into groups of 4 people.
2. Guru menyediakan dua persamaan (a) garis lurus y = 2x – 2 (b) lengkung y = x 2
The teacher prepares two equations (a) a line y = 2x – 2 (b) a curve y = x 2
3. Setiap kumpulan diberikan satu persamaan dengan arahan yang berikut:
Each group is given an equation with the following instructions:
(a) Dengan menggunakan pembezaan, cari jarak terpendek dari titik A(2, 0) kepada garis lurus atau
lengkung yang diberi.
By using differentiation, find the shortest distance from the point A(2, 0) to the given line or curve.




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02 Hybrid PBD Mate Tamb Tg5.indd 40 09/11/2021 9:24 AM

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan


(b) Dengan menggunakan kertas graf, lukis garis lurus atau lengkung yang diberi dan seterusnya, lukis
jarak terpendek dari titik A(2, 0) ke garis lurus atau lengkung.
By using the graph paper, draw the given line or curve and then, draw the shortest distance from the point A(2, 0) to the given line
or curve.
(c) Bandingkan jawapan dari (a) dan (b) dan jelaskan mana satu kaedah lebih tepat.
Compare the answers from (a) and (b) and explain which method is more accurate.
4. Bentangkan hasil dapatan dalam kelas.
Present the results in the class.
Penyelesaian: Koordinat umum pada garis lurus ialah P(x, 2x – 2) manakala koordinat umum
Solutions: pada lengkung ialah Q(x, x ). Gunakan rumus untuk mencari panjang AP atau AQ.
2
Selepas itu, gunakan d(AP) = 0 atau d(AQ) = 0 untuk mendapat nilai x yang sesuai
dx
supaya jarak ialah terpendek. dx
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The general coordinates on the line is P(x, 2x – 2) while on the curve is Q(x, x ). Use the formula to find the
2
length AP or AQ. After that, use d(AP) = 0 or d(AQ) = 0 to obtain the suitable value of x so that the distance
is the shortest. dx dx



3



7
Diberi y =  , jika x berubah daripada 5 kepada
2x −1
2
4.98. Cari
7
Given y =  , if x changes from 5 to 4.98, find (a) Perubahan hampir bagi y ialah δy.
2x −1
2
Approximate change in y is δy.

(a) perubahan hampir dalam y. (b) Nilai baharu bagi y = y asal + δy
the approximate change in y. The new value of y = y original + δy
(b) nilai hampir bagi y. (c) peratus perubahan bagi y = δy × 100%
y
the approximate value of y. δy
(c) peratus perubahan dalam y. Percentage change for y = y × 100%
percentage change in y.
7 – 1 7
(a) y = = 7(2x – 1) 2 (b) Nilai hampir y = + δy
2
 
2(5) −1
2x −1
2
2
dy = –7 (2x – 1) ( 4x) Approximate value of y
3
–
2
2
dx 2 = 1 + 0.0041 = 1.0041
–7 – 3
δy = (2x – 1) (4x)δx δy
2
2
2 (c) Peratus perubahan y = y × 100%
7 – 3 Percentage change of y
= – (2(5) – 1) ( 4(5))(–0.02) 0.0041
2
2
2 = 1 × 100%
= 0.0041
= 0.41%
E-pop Quiz
Kriteria Kejayaan: ............................................................................................ .
Saya berjaya
• Mencari kecerunan dan persamaan tangen dan normal kepada suatu lengkung yang diberi.
• Mencari titik pusingan dan menentukan ciri-ciri titik pusingan tersebut.
• Menyelesaikan soalan berkaitan dengan maksimum atau minimum.
• Mencari kadar perubahan suatu kuantiti.
• Mancari perubahan hampir.
• Menyelesaikan soalan berkaitan dengan kadar perubahan, perubahan hampir.






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02 Hybrid PBD Mate Tamb Tg5.indd 41 09/11/2021 9:24 AM

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan
Pentaksiran Sumatif
SPM





PRAKTIS PdPR Jawapan
Bab 2 Praktis PdPR Bab 2
3
Kertas 1 3. Isi padu V cm bagi suatu gas di bawah tekanan P
unit memenuhi persamaan PV = 220. Apabila P =
40, P bertambah dengan kadar 4 unit/s. Tentukan
1. Cari nilai bagi / Find the value of kadar perubahan isi padu pada ketika itu.
(a) had/lim (4 – x ) The volume V cm of a gas under the pressure P unit is given
2
3
x → 1
[2 markah / 2 marks] by the equation PV = 220. When P = 40, P increases at a rate
2
(b) f‘ (–2) jika/if f(x) = 4 – 2x + x of 4 unit/s. Determine the rate of change of the volume at that
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instant.
[3 markah / 3 marks]
[3 markah / 3 marks]
Jawapan / Answer : Jawapan / Answer :
(a) had/lim (4 – x ) = 3
2
x → 1
(b) f(x) = 4 – 2x + x PV = 220 –1
2
f‘ (x) = –2 + 2x V = 220P
f‘ (–2) = –2 + (2)(–2) = –6 dV = –220
dP P 2
dV = –220 dP
.
dt P 2 dt
2. Diberi bahawa A = 5t + 2 dan x = 2t – 6t. = –200 [4] = –0.5 cm /s
2
3
Given that A = 5t + 2 and x = 2t – 6t. 40 2
2
(a) Ungkapkan dA dalam sebutan t.
dA dx
Express in terms of t. 4. Persamaan lengkung diberi oleh y = 2x + 3.
2
dx
[2 markah / 2 marks] Tentukan perubahan hampir dalam y, dalam
(b) Cari perubahan kecil bagi x apabila A sebutan β apabila x berubah daripada β kepada
berubah daripada 4 kepada 3.98 pada ketika β + δβ.
t = 2. The equation of the curve is given by y = 2x + 3. Determine the
2
Find the small change in x when A changes from 4 to 3.98 small change in y, in terms of β when x changes from β to β +
when t = 2. δβ.
[3 markah / 3 marks] [3 markah / 3 marks]

Jawapan / Answer : Jawapan / Answer :
(a) A = 5t + 2 x = 2t – 6t y = 2x + 3 . δx = β + δβ – β
2
2
dA = 5 dx = 4t – 6 dy
dt dt dx = 4x = δβ
dA = 5 δy = dy .δx
dx 4t – 6 dx
= 4x.δβ
(b) δA = 3.98 – 4 = –0.02 = 4β.δβ
δA = dA .δt
dt
–0.02 = 5 δt
δt = –0.004 5. Dua kuantiti, x dan y berubah supaya x + y = 5.
2.
δx = dx .δt t = 2 Satu kuantiti lain z ditakrifkan oleh z = xy Cari
dt nilai-nilai x dan y yang menjadikan z maksimum.
= (4t – 6).δt Two quantities, x and y changes such that x + y = 5. Another
2
= (8 – 6)(–0.004) quantity z is defined by z = xy . Find the values of x and y which
will make z maximum.
= 2(–0.004) [7 markah / 7 marks]
= –0.008





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02 Hybrid PBD Mate Tamb Tg5.indd 42 09/11/2021 9:24 AM

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan
Jawapan / Answer : 2. 16 cm
y
x + y = 5 z = xy 2 y
2
2
= y (5 – y) = 5y – y 3
dz = 10y – 3y 2
dy = y[10 – 3y] = 0
y = 0 y = 10
3
x = 5 x = 5 Sekeping logam yang berbentuk segi empat
3 sama dengan bersempadan 16 cm, dilipat untuk
2
d z = –6y membentuk sebuah kotak terbuka setelah
dy 2 dikeluarkan 4 segi empat sama yang bersempadan
10
2

,
Apabila/When y = 10 d z = –6   = –20 , 0 y cm daripada setiap pepenjurunya.
3 dy 2 3 A square piece of metal has a side of 16 cm, is to be folded to
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5 10

∴ ,  ialah maksimum make an open box by cutting off 4 squares each with sides y cm
from each of the tips of the square.
3 3
(a) Apakah nilai y yang menjadikan kotak itu
mempunyai isi padu maksimum?
What is the value of y which will make the volume
maximum?
Kertas 2 [5 markah / 5 marks]
(b) Seterusnya, cari nilai isi padu maksimum itu.
Then, find the maximum volume.
1. Diberi bahawa persamaan suatu lengkung ialah [5 markah / 5 marks]
3
y = .
x 3
Given that the equation of a curve is y = x 3 3 . Jawapan/Answer:
2
(a) Cari nilai dy apabila x = 3. (a) V = (16 – 2y) y
dx dV
Find the value of dy when x = 3. = 2(16 – 2y)(–2)y + (16 – 2y) 2
dx dy = (16 – 2y)[–4y + 16– 2y]
[3 markah / 3 marks]
(b) Seterusnya, anggarkan nilai bagi 3 . = (16 – 2y)(16 – 6y)
3 (2.98) 3 dV
Then, estimate the value of . Apabila maksimum / minimum = 0
(2.98) 3 dy
[4 markah / 4 marks] When maximum / minimum
y = 8 atau 8
Jawapan/Answer: 3
(a) y = x 3 3 d V
2
dy = – 9 (b) dy 2 = (16 – 2y)(–6) + (16 – 6y)(–2)
dx x 4 = –2[48 – 6y + 16 – 2y]
= –2[64 – 8y]
Apabila/When x = 3 , dy = –9 = –1
2
dx 3 4 9 Apabila/When y = 8 , d V = 0
dy dy 2
(b) δy = .δx

8 d V
8
dx y = , 2 = –2 64 – 8  
1
= – [–0.02] 3 dy 2 3
9 , 0 (Maksimum)
1 8
= = ∴ y = cm.
900 450 3

2 8
8
3 1 V = 16 – 2     = 303 11 cm 2
∴ y + δy = + 3 3 27
27 450
1
= + 1 = 17 = 0.113
9 450 150







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02 Hybrid PBD Mate Tamb Tg5.indd 43 09/11/2021 9:24 AM

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan











1. A 8 m D
x m
F G





B E x m C
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Rajah menunjukkan sebuah dinding segi empat sama ABCD dengan Ungkapkan luas pintu
ukuran 8 m × 8 m. CEFG ialah pintu. Bahagian dinding akan dicatkan. Diberi dalam sebutan x. Untuk luas
d L
2
EC = GD = x m. KBAT Menganalisis minimum, dx 2 . 0.
The diagram shows a square wall ABCD with a dimension of 8 m × 8 m. CEFG is a door. The shaded Express the area of door in terms
2
region is to be painted. Given EC = GD = x m. of x. For minimum area, d L 2 . 0.
dx
(a) Tunjukkan bahawa dinding akan dicat mempunyai luas L = 64 – 8x + x 2.
Show that the wall to be painted has an area of L = 64 – 8x + x .
2
(b) Cari nilai x supaya luas dinding dicat adalah minimum.
Find the value of x so that the wall to be painted is minimum.
(c) Apakah harga perbelanjaan mengecat jika kos bagi satu meter persegi
ialah RM3.50.
What is the cost of painting the wall if it costs RM35 for one square metre.
(a) L = 8 – x(8 – x) (c) L = 64 – 8(4) + 4 = 48 m 2
2
2
= 64 – 8x + x 2 Kos/Cost = 48 × RM3.5 = RM168

(b) dL = –8 + 2x
dx
Untuk maksimum/minimum
dL = 0 ; x = 4 m
dx
2
d L = 2 . 0 ∴ minimum Sudut dicangkum ialah 90°,
1
dx 2 maka luas sektor = pj .
2
4
Untuk luas maksimum,
2. Rajah menunjukkan dua buah bulatan sepusat O. Jejari d L 2 , 0.
2
untuk bulatan besar ialah 8 cm. Perbezaan jejari di antara dx
bulatan besar dan kecil ialah 2x cm. The subtended angle is 90°, hence
1
2
The diagram shows two circles with the same centre O. The radius of the O 2x the area of sector = pj . For
4
2
big circle is 8 cm. The difference between the radius of the big circle and the 8 cm maximum area, d L 2 , 0 .
small circle is 2x cm. KBAT Mengaplikasi dx
(a) Tunjukkan bahawa luas rantau berlorek L ialah
L = p(8x – x ).
2
Show that the area of the shaded region L is L = p(8x – x ).
2
(b) Seterusnya, cari nilai x supaya luas maksimum tercapai dan apakah luas
maksimum itu?
Then, find the value of x so that the maximum area is obtained and what is the maximum area?
1
2
(a) L = p[8 – (8 – 2x) ] (b) dL = p[8 – 2x] = 0
2
4 dx x = 4 cm
1 d L
2
2
= p[32x – 4x ] = p[–2] , 0 ∴ maksimum / maximum
4 dx 2
= p[8x – x ] L = p[8(4) – 4 ]
2
2
= 16 p cm 2 Kuiz 2

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