Add Math form 5 Chapter 2 Full Solution

Apabila y = 3, 3 = ! 2x + 7

9 = 2x + 7

2x = 2

x=1

dan dy = 1 + 7
dx ! 2(1)

= 1
! 9

= 1
3
dy dy dx
dt = dx × dt

6 = 1 × dx
= 3 dt
dx 18
dt unit s–1

3. (a) y = (x – 8)! x + 4

[ ]dy = (x – 8) 1 (x + 4)– 12 (1) + ! x + 4(1)
2
dx
x–8 + ! x + 4
= 2! x + 4

= x – 8 + 2(x + 4)
2! x + 4

= x –8+ 2x + 8
2! x +4

= 3x
2! x + 4

(b) Apabila x = 5, dy = 3(5)
dx 2! 5 + 4

= 15
2(3)

= 15
6
5
= 2

dy = dy × dx
dt dx dt
ddyt =
5 ×6
2

= 15 unit s–1

Perbincangan (Halaman 68)
1
(a) V = 3  π r2h …1

8r = h
16
h
r = 16 × 8

= h …2
2

Gantikan 2 ke dalam 1:

( )V =1  π  h 2h
3 2

= 1  π h3
12

51

Kadar perubahan V diberi oleh:

dV = dV × dh (petua rantai)
dt dh dt

( )=ddh  1 πh3 × dh
12 dt

= 1  πh2 × dh
4 dt
dV
Apabila h = 8 dan dt = 64π, kita peroleh

64π = 1  π(8)2 × dh
4 dt
dh
64π = 16π × dt

dh = 4
dt

Jadi, kadar perubahan kedalaman air dalam bekas itu ialah 4 cms–1.
(b) L = πr2 …3
Gantikan 2 ke dalam 3:
( )L h 2
= π  2

= 1  πh2
4

Kadar perubahan L diberi oleh:

dL = dL × dh (petua rantai)
dt dh dt

( )=d 1 πh2 × dh
dh 4 dt

= 1  πh × 4
2
dh
Apabila h = 8 dan dt = 4, kita peroleh

dL = 1  π(8) × 4
dt 2

= 16π

Jadi, kadar perubahan luas permukaan mengufuk ialah 16π cm2s–1.

Latihan Kendiri 2.13

1. y = 1  x2
8
dy
dx = 1  x
4
dy 1
Apabila x = 4, dx = 4 (4) = 1

dy = dy × dx
dt dx dt
dy
dt = 1 × 3

= 3 unit s–1

2. L = x2
dL
dx = 2x

Apabila L = 4,

x2 = 4
x2 = ! 4

dL = = 2 (. 0)
dan dx 2(2) = 4

52

dL = dL × dx
dt dx dt
dx
8 = 4 × dt

ddxt = 8
4

= 2 cms–1

3. I = x3
dI
dx = 3x2

Apabila x = 10, dI = 3(10)2
dx
= 300
dI dI dx
dt = dx × dt

–10.5 = 300 × dx
dt

dx = – 1300.05 = – 2070 cmmin–1
dt

4. (a) V = π r2h

= π(3)2h

= 9πh
dV
(b) dt = 9π × – 0.6

= –5.4π cm3min–1

5. Katakan panjang bayang-bayang dan panjang hujung bayang-bayang dari kaki tiang lampu masing-masing ialah

s m dan l m.
6 1.8
x+s = s

6s = 1.8(x + s)

6s = 1.8x + 1.8s 6m
4.2s = 1.8x
3
s = 7 x 1.8 m

(a) s= 3 x xm sm
7 lm
ds 3
dx = 7

Jadi, ds = ds × dx
dt dx dt

= 3 × 3.5
7

= 1.5

Maka, kadar perubahan panjang bayang-bayang ialah 1.5 ms–1.

(b) l = x + s
3
= x + 7 x

= 170 x
dl 10
dx = 7

Jadi, dl = dl × dx
dt dx dt

= 10 × 3.5
7

= 5

Maka, kadar perubahan hujung bayang-bayang yang bergerak ialah 5 ms–1.

Perbincangan (Halaman 71)
Boleh digunakan tetapi jawapannya tidak tepat dan bukan penghampiran yang terbaik.

53

Latihan Kendiri 2.14

1. (a) y = 4x3 – 3x2

dy = 12x2 – 6x
dx

Apabila x = 1, dx = 1.05 – 1

= 0.05
dy
dan dx = 12(1)2 – 6(1)
= 12 – 6

=6

Jadi, dy ≈ dy × dx
dx

= 6 × 0.05

= 0.3 unit
(b) y = 4! x + 3x2

( )dy = 4 1 x– 12 + 6x
2
dx
2
= ! x + 6x

Apabila x = 4, dx = 3.98 – 4

= – 0.02

dan dy = 2 + 6(4)
dx ! 4

= 1 + 24

= 25

Jadi, dy ≈ dy × dx
dx

= 25 × – 0.02

= – 0.5 unit
3
2. (a) y = 2x2
dy 1
dx =
3x2

= 3! x 3

Apabila y = 16, 16 = 2x2

3

x2 = 8

( ) x= 23 2
3

=4

dy = 15.7 – 16

= – 0.3

dan dy = 3! 4
dx
= 3(2)

=6
dy
Jadi, dy ≈ dx × dx

– 0.3 = 6 × dx

dx = – 06.3
= – 0.05 unit
x + 2
(b) y = 2

= 1 x + 1
2
dy
dx = 1
2

54

Apabila y = 2, 2 = x + 2
2

4=x+2

x=2

dy = 2 + p – 2

= p

Jadi, dy ≈ dy × dx
dx
1
p= 2 × dx

dx = 2p unit
16
3. y = x2 = 16x–2

dy = –32x–3 = – 3x23
dx
16
Apabila x = 2, y = 22

y = 16
4
=4

dx = 2.02 – 2

= 0.02
dy – 3223
dan dx =

= – 4 dy
dx
f (x + dx) ≈ y +  dx

(2 16 = 4 + (– 4)(0.02)
+ 0.02)2
= 4 – 0.08

= 3.92

5
4. y = x4
dy 5 1
dx = 4 x 4

dy = 4 y
100
= 0.04y
5
= 0.04x4

dy ≈ dy × dx
dx

5 = 5 1 × dx
4
0.04x4 x4

5
0.04x4
dx = 1
5
4 x4

= 0.032x

Peratus perubahan hampir dalam x = dx × 100%
x

= 0.032x × 100%
x

= 3.2%

55

Latihan Kendiri 2.15
!  1.
T = 2π  l
10
l – 21
[ ( ) ( )] ddTl 1 10 1
= 2π 2 10

!  = π

10 l
10
dT π
! Apabila l = 9, dl =
9
10 10

= π! 10
30

dan dl = 9.05 – 9 = 0.05

Maka, dT ≈ dT × dl
dl
π! 10
= 30 × 0.05

= π! 10 saat
600
2. L = π j2

dL = 2π j
dj
L = 4π

π j2 = 4π

j2 = 4

j = 2 (. 0)

Apabila j = 2, ddLj = 2π(2)
= 4π

dL = 4.01π – 4π
= 0.01π

dan dL ≈ dL × dj
dj

0.01π = 4π × dj

dj = 0.01π
4π

= 0.0025 cm

3. V = x3

dV = 3x2
dx

Apabila x = 2, dx = 1.99 – 2

dV = – 0.01
dx
dan = 3(2)2
= 12

Maka, dV = dV × dx
dx

= 12 × – 0.01

= – 0.12 cm3
4
4. I = 3 π j3

dI = 4π j2
dj

Apabila j = 5, dj = 4.98 – 5

= – 0.02

56

dan dI = 4π (5)2
dj = 100π

Maka, dI ≈ dI × dj
dj

= 100π × – 0.02

= –2π cm3

Latihan Formatif 2.4

1. (a) y = ! x + 1

dy = 1 (x + 1)– 12 (1)
dx 2
1
= 2! x + 1

Apabila x = 0, ddyx = 1
2! 0 + 1

= 1
2
1
Persamaan tangen: y–1= 2 (x – 0)

2y – 2 = x
2y – x = 2

Pada paksi-x, y = 0

2(0) – x = 2

x = –2

º Q(–2, 0)

(b) Persamaan normal: y – 1 = –2(x – 0)

y – 1 = –2x

y = –2x + 1

Pada paksi-x, y = 0

0 = –2x + 1

2x = 1
1
( ) 1 x= 2
2
ºR , 0

(c) Luas ∆ PQR = 1   0 –2 1 0
2 1 0 2 1
0

  =1 1
2 2 – (–2) 

= 1  21 + 2 
2

= 5
4
1
= 1 4 unit2

2. (a) 2y = 4 – x 1
2
y = 2 – x

y = x2 – 4x + 1

dy = 2x – 4
dx

Jadi, 2x – 4 = 2
2x = 2 + 4
2x = 6

x=3

57

Apabila x = 3, y = 32 – 4(3) + 1

= 9 – 12 + 1

= –2

º a = 3, b = –2

(b) Persamaan tangen: y + 2 = 2(x – 3)

y + 2 = 2x – 6

y = 2x – 8

Pada paksi-x, y = 0

0 = 2x – 8

2x = 8

x=4

º B(4, 0)

(c) Persamaan normal: y + 2 = – 12 (x – 3)
2y + 4 = –x + 3

2y + x + 1 = 0

Pada paksi-x, y = 0

2(0) + x + 1 = 0

x = –1

º C(–1, 0)

(d) Luas ∆ BPC = 1   4 –1 3 4
2 0 0 –2 0

 = 1  2 – (–8) 
2
 10 
= 1
2

= 5 unit2

3. (a) Luas = 75

x2 + 4hx = 75

4hx = 75 – x2

h = 75 – x2
4x

Isipadu, V = x2h
( ) = x2
1 75 – x2
4 4x

= x(75 – x2)

= 1 (75x – x3) (Tertunjuk)
4
(b) Untuk V maksimum, ddVx = 0

75 – 3 x2 = 0
4 4
75 3
4 = 4 x2

3x2 = 75

x2 = 25
x = 5 (. 0)
1
4
V = [75(5) – 53]

= 1 (375 – 125)
4
1
= 4 (250)

= 62.5 cm3

d 2V = – 32 x
dx2

58

Apabila x = 5, d 2V = – 23 (5)
dx2
= –7.5 (, 0) Ú V adalah maksimum

Maka, V mempunyai nilai maksimum apabila x = 5 cm dan isi padu maksimum kotak ialah 62.5 cm3.

4. (a) x2 + y2 = 102

y2 = 100 – x2
y = ! 100 – x2

dy = 1 (100 – x2)– 21 (–2x)
dx 2
x
= –  ! 100 – x2

dy = dy × dx
dt dx dt

= –  x – x2 ×3
! 100
3x
= –  ! 100 – x2

Apabila x = 8, dy = –  3(8) 82
dt ! 100 –

= – 4

Jadi, kadar perubahan hujung kayu A ialah – 4 ms–1.

(b) x2 + y2 = 102

Apabila y = 6, x2 + 62 = 102

x2 = 100 – 36

x = ! 64

= 8
dy 8
Apabila x = 8, dx = –  ! 100 – 82

= – 43
dy dy dx
dan dt = dx × dt

–2 = – 43 × dx
dt
dx
dt = 1.5

Jadi, kadar perubahan hujung kayu B ialah 1.5 ms–1.

5. Katakan x m ialah jarak mengufuk antara helikopter dengan budak lelaki dan z m ialah jarak antara helikopter

dengan budak lelaki pada masa t.

z2 = x2 + 1352 xm
z = ! x2 + 18 225

dz = 1 (x2 + 18 225)– 21(2x)
dx 2 x
=
! x2 + 18 225 135 m zm

Jadi, dz = dz × dx
dt dx dt
dx ddzt 72
Apabila x = 72 dan dt = –17, = ! 722 + 18 225 × –17

= 72 × –17
153

= –8

º –8 ms–1

59

Latihan Sumatif
( ) [ ] 1. (a)
had   8 + 2x – x2 = had (4 – x)(2 + x)
8 – 2x2 2(4 – x2)
x ˜ –2 x ˜ –2

[ ]=
had (4 – x)(2 + x)
2(2 + x)(2 – x)
x ˜ –2

[ ]= had
x ˜ –2
(4 – x)
2(2 – x)

= 4 – (–2)
2[2 – (–2)]

= 6
8

= 3
4
(b) had

( ) ( )( )x˜0
! 1 +x+ x2 –1 = had ! 1 + x + x2 – 1 ! 1 + x + x2 + 1
x x ! 1 + x + x2 + 1
x˜0

[ ]( )= had
x˜0
1 + x + x2 – 1
x ! 1 + x + x2 + 1

[ ]( )= had
x˜0
x + x2
x ! 1 + x + x2 + 1

[ ]( )= had
x˜0
x(1 + x)
x ! 1 + x + x2 + 1

( )= had
x˜0
1+x
! 1 + x + x2 + 1

= 1 1
! 1 +

= 1
2

(c) had 9 – x 2 =8
4 – ! x 2 + 7
x˜k

9 – k 2 =8
4 – ! k 2 + 7
( )( )
9 – k 2 4 + ! k 2 + 7 =8
4 – ! k 2 + 7 4 + ! k 2 + 7
( )(9 – k 2) 4 + ! k 2 + 7
=8
16 – (k2 + 7)
( )(9 – k 2) 4 + ! k 2 + 7
=8
9 – k2

4 + ! k 2 + 7 = 8
! k 2 + 7 = 4
k2 + 7 = 16

k2 = 9

k = ±3

( ) 2. hada–5 = –3
x ˜ –1x+4
a–5
–1 + 4 = –3

a–5 = –3
3
a – 5 = –9

a = –9 + 5

a = – 4

60

( ) 3. d d
(a) dx 1 = dx [(2x + 1)–1]
2x + 1

= –1(2x + 1)–2(2)

= – (2x 2 1)2
+

( b) ddx[4x(2x – 1)5] = 4x[5(2x – 1)4(2)] + (2x – 1)5(4)
= 40x(2x – 1)4 + 4(2x – 1)5

= 4(2x – 1)4[10x + (2x – 1)]

= 4(12x – 1)(2x – 1)4
[ ](c)
d (2 6 = d [6(2 – x)–2]
dx – x)2 dx

= 6(–2)(2 – x)–3(–1)

= 12
(2 – x)3
( )(d) + 3)– 21(1) + ! x +
d (x! x + 3) = x 1 (x 3(1)
dx 2
x
= 2! x + 3 + ! x + 3

= x + 2(x + 3)
2! x + 3

= 3x + 6
2! x + 3

= 3(x + 2)
2! x + 3

4. (a) y = x(3 – x)

= 3x – x2
ddxy = 3 – 2x
d 2y
dx2 = –2

y dd x2y2 + x ddyx + 12 = (3x – x2)(–2) + x(3 – 2x) + 12
= –6x + 2x2 + 3x – 2x2 + 12

= 12 – 3x

(b) 12 – 3x = 0

3x = 12

x = 12
3

x = 4

5. y = ax + b
x2
= ax + bx–2

ddxy = a – 2bx–3
2b
= a – x3

( )Pada titik , – 72 b
–1, – 27 = a(–1) + (–1)2

– 27 = –a + b
7
a–b= 2

2a – 2b = 7 …1

61

dan dy = 2
dx
2b
a – (–1)3 = 2

a + 2b = 2 …2

1 + 2: 3a = 9
a=3

Gantikan a = 3 ke dalam 1: 2(3) – 2b = 7

6 – 2b = 7

2b = –1

– 21 b = – 12

ºa = 3, b =
V= 4 π r3
6. 3

dV = 4π r2
dr

dV = dV × dr
dt dr dt

20π = 4π r2 × 0.2

20 = 0.8r2

r2 = 25

r = 5 (. 0)

º r = 5 cm = 14(6x3 + 1)– 21

7. (a) y = 14
! 6x3 + 1

( ) dy – 21 (6x3 + 1)– 23(18x2)
dx = 14

= –  126x2
3
(6x3 + 1)2

Apabila x = 2, dx = 2.05 – 2

= 0.05

dan dy = –  126(2)2
dx 3

[6(2)3 + 1]2

= – 350443

= – 4729

Jadi, dy ≈ dy × dx
dx
= – 4729 × 0.05

= –0.0735 unit
14
(b) Apabila x = 2, y = ! 6(2)3 + 1

= 14
7

= 2

dx = 2.05 – 2 = 0.05

dan dy = – 7429
dx

14
! 6(2.05)3 + 1
( ) – 7492 (0.05)
=2+

= 2 – 0.0735

= 1.927

62

8. y = 1
! x

= x– 12

dy = – 21 x– 23
dx

= – 2!1 x3

Apabila x = 4, y = 1
! 4

= 1
2

dx = 2 × 4 = 0.08
100
dy
dan dx = – 2!1 43 = – 116

Jadi, dy ≈ dy × dx
dx
= – 116 × 0.08

= – 0.005

Maka, peratus perubahan hampir dalam y = dy × 100%
y

= – 0.0105 × 100%
2

= –1%

9. y = 3x2 – 4x + 6

dy = 6x – 4
dx

Apabila x = 2, y = 3(2)2 – 4(2) + 6

= 12 – 8 + 6

dy = 10
dx
= 6(2) – 4

=8

dan dx = p × 2
100
= 0.02p

dy = dy × dx
dx

= 8 × 0.02p

= 0.16p

Maka, peratus perubahan dalam y ialah 0.16p × 100 = 1.6p%
10
10. (a) Titik maksimum ialah (–1, 6) dan titik minimum ialah (1, 2)

(b)

y

(–1, 6) y = f (x)

(1, 2)
0x


63

11. (a) y = 3x3 – 4x + 2

dy = 9x2 – 4
dx

Pada titik A(2, 1), dy = 9(2)2 – 4
dx
= 36 – 4

= 32

Persamaan tangen di titik A(2, 1) ialah:

y – 1 = 32(x – 2)

y – 1 = 32x – 64

dy y = 32x – 63
dx
(b) = 32

9x2 – 4 = 32

9x2 = 36

x2 = 4

x = ±2

Apabila x = –2, y = 3(–2)3 – 4(–2) + 2

= –24 + 8 + 2

= –14

º (–2, –14)

( ) 12. (a) j2 = 6! 3 2 – t2 A
t cm
j = ! 108 – t2 …1 D

I = 1  πj2t …2
3
Gantikan 1 ke dalam 2: 6� 3 cm
B j cm
I = 1  π (! 108 – )t2 2t
3
1
= 3  π (108 – t2)t

= 36π t – 1  π t3 ddIt  
3
Untuk isi padu maksimum, = 0

36π – π t2 = 0

36π = πt2

36 = t2
t = 6 (. 0)


º t = 6 cm

(b) I = 36π t – 1  π t3
3
1
Apabila t = 6, I = 36π(6) – 3  π (6)3

= 216π – 72π

= 144π

º Isi padu kon ialah 144π cm3.

13. AC = ! 302 + x2 = ! 900 + x2

Jumlah masa yang diambil dari A ke D ialah

T= ! 900 + x2 + 400 – x
40 50

T = 1 ! 900 + x2 + 8 – 1  x
40 50

64

Untuk nilai pegun T, dT  = 0
dx
(900 + x2)– 12(2x) –
( )11 1 = 0
2 50
40 x 1
40! 900 + x2 50
=

50x = 40! 900 + x2

5x = 4! 900 + x2

25x2 = 16(900 + x2)

25x2 = 14 400 + 16x2

9x2 = 14 400

x2 = 1 600

x = ! 1 600

= 40

º Jarak dari B ke C ialah 40 m.

14. I = 8

x3 = 8

x = 2

L = 6x2
AddLxpab=il1a2xx
= 2, dL   = 12(2)
dx = 24

dL = dL × dx
dt dx dt
dL
dt = 24 × 2

= 48

º Kadar perubahan jumlah luas permukaan kubus ialah 48 cm2s–1.

15. (a) Luas, A = 1 xy
2
1
= 2 x(6x – x2)

= 1 (6x2 – x3) (Tertunjuk)
2

(b) (i) A= 1 (6x2 – x3) = 3x2 – 1 x3
2 2
dA 3
dx = 6x – 2 x2

dA = dA × dx
dt dx dt
( ) dA 3
dt = 6x – 2 x2 (2)

[ ] Apabila = 2, dA = 3 (2)2 (2)
x dt 6(2) – 2

= (12 – 6)(2)

= 12

Kadar tokokan A ialah 12 unit2 s–1

[ ](ii) Apabila x = 5, dA = 3 (5)2 (2)
dt 6(5) – 2

= (30 – 37.5)(2)

= –15

º Kadar susutan A ialah 15 unit2 s–1

65

16. (a) r = h
12 20

r = 12  h
20

r = 3  h …1
5

V = 1 π r2h …2
3
( )Gantikan
1 ke dalam 2: V = 1  π  3  h 2h
3 5
( ) 1 9
= 3  π  25 h2 h

= 3  π h3 (Tertunjuk)
3 25
(b) (i) V= 25  π h3

dV = 9  π h2
dh 25

Apabila h = 5, dh = 4.99 – 5 = – 0.01

dan ddVh = 9  π (5)2
25

= 9π

d V ≈ dV × d h
dh

d V = 9π × – 0.01

= – 0.09π

º Perubahan kecil dalam isi padu air ialah – 0.09π cm3.

(ii) Jika h menyusut sebanyak p%,
d h = – 1p00(h) = – 1p0h0
dV
d V ≈ dh × d h

d V = 9  π h2 × – 1p0h0
25
9
= – 2 500 π ph3

Jadi, dV × 100% = – 2 9 π ph3 × 100%
V 500
3
25  π h3

= –3p

º Isi padu menyusut sebanyak 3p%.

66