Solution Manual - Textbook Add Math BM F5 KSSM

Jawapan

BAB 1  SUKATAN MEMBULAT

Aktiviti Penerokaan 1 (Halaman 2)

3. Satu radian ialah sudut yang tercangkum pada pusat bulatan oleh lengkok yang sama panjang dengan jejari bulatan.

π rad = 180°

4. 1 rad = 180° ≈ 57.29° dan 1° = π ≈ 0.01746 rad
π 180°

Perbincangan (Halaman 3)

Jawapan yang diperoleh lebih jitu dan tepat terutamanya dalam mengukur panjang lengkok atau menentukan perimeter

tembereng suatu objek yang berbentuk bulatan.

Latihan Kendiri 1.1

1. (a) π rad = π × 180°
8 8 π

=81 × 180°

= 22.5°

(b) 3  π rad = 3 π × 180°
4 4 π

=43 × 180°

= 135°

(c) 0.5 rad = 0.5 × 180°
3.142

= 28° 39'

(d) 1.04 rad = 1.04 × 180°
3.142

= 59° 35'

2. (a) 18° = 18° × π
180°
=110  π rad

(b) 120° = 120° × π
180°
=23  π rad

(c) 225° = 225° × π
180°
1
= 1 4  π rad

(d) 300° = 300° × π
180°
2
= 1 3  π rad

Latihan Formatif 1.1

1. (a) 172 π rad = 172 π × 180°
π

=172 × 180°

= 105°

1

(b) 1 1  π rad = 4 π × 180°
3 3 π

=34 × 180°

= 240°

(c) 2 rad = 2 × 180°
3.142

= 114° 35'

(d) 4.8 rad = 4.8 × 180°
3.142

= 274° 59'

2. (a) 76° = 76° × 3.142
180°

= 1.327 rad

(b) 139° = 139° × 3.142
180°

= 2.426 rad

(c) 202.5° = 202.5° × 3.142
180°

= 3.535 rad

(d) 320° 10' = 320° 10' × 3.142
180°

= 5.589 rad

3. (a) 73° = 73° × 3.142
180°

= 1.274 rad

(b) 118° = 118° × 3.142
180°

=2.060 rad

(c) 150.5° = 150.5° × 3.142
180°

= 2.627 rad

(d) 220° = 220° × 3.142
180°

= 3.840 rad

Aktiviti Penerokaan 2 (Halaman 5)

4. Nilai kedua-duanya adalah sama.

5. Masih sama.

6. Jika ∠AOB = q diukur dalam radian, jadi

Panjang lengkok minor AB = Lilitan bulatan
q 2π

s = 2πj
q 2π
2πj
s= 2π ×q

s = jq

Perbincangan (Halaman 6)

Saiz sudut, q diberi oleh nisbah panjang lengkok, s kepada panjang jejari bulatan, j, iaitu:

q= s
j

Jadi, s = jq, dengan q ialah sudut dalam radian.

2

Latihan Kendiri 1.2

1. (a) Panjang lengkok MN = jq

= 12(1.1)

= 13.2 cm

(b) Panjang lengkok MN = jq

= 8(2)

= 16 cm

(c) Panjang lengkok MN = jq

= 5( 5 π)
6
= 13.09 cm

(d) Panjang lengkok MN = jq

= 10(3.142 – 2.45)

= 10(0.692)

= 6.92 cm

2. (a) sEF = 25

j(6.284 – 1.284) = 25

5j = 25

j = 25
5
= 5 cm

(b) sEF = jq
= 5(1.284)
= 6.42 cm

3. (a) sQR = 5.7

5q ' = 5.7

q ' = 5.7
5
= 1.14 rad

º q = 3.142 – 1.14

= 2.002 rad

(b) sPQ = jq
= 5(2.002)

= 10.01 cm

Kuiz Pantas (Halaman 8)

Boleh, iaitu: AC 10
sin 114° sin 33°
=

AC = 10 × sin 114°
sin 33°

= 16.77 cm

Latihan Kendiri 1.3

1. (a) sABC = 6(2.5)
= 15 cm

2.5 rad = 2.5 × 180°
3.142

= 143° 13'

AC2 = 62 + 62 – 2(6)(6)(kos 143° 13')

AC = ! 129.6652

= 11.39 cm

º Perimeter tembereng berlorek ABC = 15 + 11.39

= 26.39 cm

3

( )(b) sABC = 10 π
3

= 10.47 cm

AC2 = 102 + 102 – 2(10)(10)(kos 60°)

= 100

AC = ! 100

= 10 cm

º Perimeter tembereng berlorek ABC = 10.47 + 10

3.142 = 20.47 cm
180°
(c) 120° = 120° ×

= 2.095 rad

SABC = 8(2.095)
= 16.76 cm

AC2 = 82 + 82 – 2(8)(8)(kos 120°)

= 192

AC = ! 192

= 13.86 cm

º Perimeter tembereng berlorek ABC = 16.76 + 13.86

= 30.62 cm

(d) sin ˙AOD = 7.5
9
O
˙AOD = 56° 27'
3.142
˙AOC = 2(56° 27') × 180° 9 cm

= 112° 54' × 3.142 A 7.5 cm D
180°
= 1.971 rad

sABC = 9(1.971)
= 17.739 cm

AC2 = 92 + 92 – 2(9)(9)(kos 112° 54')

= 225.0381

AC = ! 225.0381

= 15.001 cm

º Perimeter tembereng berlorek ABC = 17.739 + 15.001

= 32.74 cm

2. (a) sPQ = 14

7q = 14

q = 14
7

= 2 rad

2 rad = 2 × 180°
3.142

= 114° 35'

(b) PQ2 = 72 + 72 – 2(7)(7)(kos 114° 35')

= 138.7696

PQ = ! 138.7696

= 11.78 cm

º Perimeter tembereng berlorek = 11.78 + 14

= 25.78 cm

Latihan Kendiri 1.4

1. (a) 110° = 110° × 3.142
180°

= 1.92 rad

4

sAB = 4(1.92)
= 7.68 cm

sCD = 9(1.92)
= 17.28 cm

º Perimeter kawasan berlorek = 2(5) + 7.68 + 17.28

= 34.96 cm

(b) sCD = 3
4q = 3
3
q = 4 rad

( ) sAB = 3 3
4

= 2.25 cm

( ) sCD = 4 3
= 3 cm 4

º Perimeter kawasan berlorek = 2(1) + 2.25 + 3

π = 7.25 cm
2
(c) ˙AOB =π – – 0.5

= 1.071 rad

0.5 rad = 0.5 × 180°
3.142

= 28° 39'
10
AC = tan 28° 39' O
10 cm
AC = tan 10 39'
28° A

= 18.303 cm

OC = ! 102 + 18.3032 28° 39’ C

= 20.857 cm

BC = 20.857 – 10
= 10.857 cm

sAB = 10(1.071)
= 10.71 cm

º Perimeter kawasan berlorek = 10.71 + 18.303 + 10.857
= 39.87 cm

2. Sudut antara Washington dan Lima == 5(308.9.828°°×+311.81204.0°24°) × 31.8104°2


= 0.889 rad

Jarak antara Washington dan Lima = 6 371(0.889)

= 5 663.819 km

3. 85° = 85° × 3.142
180°

= 1.484 rad

Jarak = 25(1.484)
= 37.1 m

4. (a) s = 35(π)
= 35(3.142)
= 109.97 cm
(b) Perimeter = 70 + 2(100) + 109.97
= 379.97 cm

5

5. 2πj = 50.8
50.8
j = 2π

=2(35.01.842)

= 8.084 cm

185° = 185° × 3.142
180°

= 3.229 rad

Panjang lengkok = 8.084(3.229)

= 26.103 cm

2πj = 30.5
30.5
j = 2π

=2(33.01.542)

= 4.854 cm

160° = 160° × 3.142
180°

= 2.793 rad

Panjang lengkok = 4.854(2.793)

= 13.557

º Panjang rantai = 2(25) + 26.103 + 13.557

= 89.66 cm

Latihan Formatif 1.2

1. (a) ˙ROS = (360° – 275°) × 3.142
=85° × 3.142 180°
180°

= 1.484 rad

(b) sRS = 15

j(1.484) = 15
15
j = 1.484

= 10.11 cm

2. sUV = 5
jq = 5 …1

Perimeter sektor UOV = 18 cm
2j + jq = 18 …2
Gantikan 1 ke dalam 2: 2j + 5 = 18

2j = 13

j = 13
2

= 6.5

Gantikan j = 6.5 cm ke dalam 1: 6.5q = 5

q = 0.7692 rad

3. (a) kos q = 4
5

q = 36° 52'

= 36° 52' × 3.142
180°

= 0.6435 rad

(b) EG = ! 52 – 42

=!  9

= 3 cm

6

sEF = 5(0.6435)
= 3.218 cm

º Perimeter kawasan berlorek = (5 – 4) + 3 + 3.218
= 7.218 cm

4. (a) Perimeter kawasan berlorek = 18 cm

2h(0.5) + 2h + 3h(0.5) = 18

4.5h = 18

h = 18
4.5

= 4 cm

(b) sRS = 12(0.5)
= 6 cm

sPQ = 8(0.5)
= 4 cm

º Beza panjang lengkok = 6 – 4
= 2 cm

5. (a) 51° = 51° × 3.142
180°
= 0.8902 rad

sMN = 10(0.8902)
= 8.902 cm

(b) 51° = 25° 30' M
2 P
M10P = tan 25° 30' 10 cm
O 25°30’
MP = 10(tan 25° 30')

= 4.770 cm

º Perimeter kawasan berlorek = 2(4.770) + 8.902
= 18.44 cm

6. s = q

3.142
180°
=( ) 36 21° ×

= 13.1964 cm

º Jumlah jarak = 2(13.1964)
= 26.39 cm

7. Lilitan tayar = 2πj
= 2(3.142)(33)
= 207.372 cm
(a) 50 pusingan lengkap = 50 × 207.372
= 10 368.6 cm
= 103.686 m
(b) 1 000 pusingan lengkap = 1 000 × 207.372
= 207 372 cm
= 2 073.72 m

Aktiviti Penerokaan 3 (Halaman 12)
4. Nilai kedua-dua nisbah adalah sama.
5. Masih sama.

7

6. Jika ∠AOB = q diukur dalam radian, jadi

Luas sektor minor AOB = Luas bulatan
q 2π

L = πj 2
q 2π
πj 2
L = 2π ×q

L = 1  j 2q
2

Latihan Kendiri 1.5

1. (a) Luas sektor AOB = 1 (62)(1.1)
2

= 19.8 cm2

(b) Luas sektor AOB = 1 (102)(2.15)
2

= 107.5 cm2

(c) ˙AOB = 2π – 5 π
3

= 1.047 rad

Luas sektor AOB = 1 (52)(1.047)
2

= 13.09 cm2

(d) 135° = 135° × 3.142
180°

= 2.357 rad

Luas sektor AOB = 1 (202)(2.357)
2

= 471.4 cm2

2. Panjang lengkok, s = 6 cm

5q = 6
6
q = 5

= 1.2 rad
1
Luas, L = 2  j2q

= 1 (5)2(1.2)
2
=21 (25)(1.2)

= 15 cm2

3. (a) Luas = 195 cm2

21  j2(3.9) = 195

1.95j2 = 195

j2 = 195
1.95

= 100

j = ! 100

= 10 cm

(b) sEF = jq
= 10(3.9)

= 39 cm

(c) Perimeter sektor major EOF = 2(10) + 39

= 59 cm

8

4. (a) Luas sektor VOW = 60 cm2

1 (102)q = 60
2
60
50q = 50

= 1.2 rad

(b) sVW = 10(1.2)
= 12 cm

(c) Perimeter sektor VOW = 2(10) + 12

= 32 cm

Latihan Kendiri 1.6

1. (a) 1.5 rad = 1.5 × 180°
3.142

= 85° 56'

Luas sektor AOB = 1 (72)(1.5)
2

= 36.75 cm2

Luas ∆AOB = 1 (7)(7)(sin 85° 56')
2

= 24.44 cm2

º Luas tembereng ACB = 36.75 – 24.44

= 12.31 cm2

( )(b) Luas sektor AOB = 1 2
2 (102) 3 π

= 104.73 cm2

Luas ∆AOB = 1 (10)(10)(sin 120°)
2

= 43.30 cm2

º Luas tembereng ACB = 104.73 – 43.30
= 61.43 cm2

(c) 58° = 58° × 3.142
180°

= 1.012 rad

Luas sektor AOB = 1 (52)(1.012)
2

= 12.65 cm2

Luas ∆AOB = 1 (5)(5)(sin 58°)
2

= 10.601 cm2

º Luas tembereng ACB = 12.65 – 10.601

= 2.049 cm2

(d) sin q = 7.5 A
9

q = 56° 27' 3.142
180°
˙AOB = 2(56° 27') ×

= 112° 54' × 3.142 7.5 cm 9 cm
180° D θ
O
= 1.971 rad

Luas sektor AOB = 1 (92)(1.971)
2

= 79.83 cm2

Luas ∆AOB = 1 (9)(9)(sin 112° 54')
2

= 37.31 cm2

9

º Luas tembereng ACB = 79.83 – 37.31
= 42.52 cm2

2. (a) sMN = 5 cm

3q = 5

q = 5
3

= 1.667 rad

1.667 rad = 1.667 × 180°
3.142

= 95° 30'

º ˙MON = 95° 30'

(b) Luas sektor MON = 1 (32)(1.667)
2

= 7.502 cm2

Luas ∆MON = 1 (3)(3)(sin 95° 30')
2

= 4.479 cm2

º Luas tembereng = 7.502 – 4.479
= 3.023 cm2

3. (a) 60° = 60° × 3.142
180°

= 1.047 rad

(b) Luas sektor HOK = 1 (42)(1.047)
2

= 8.376 cm2

Luas ∆HOK = 1 (4)(4)(sin 60°)
2

= 6.928 cm2

º Luas tembereng = 8.376 – 6.928
= 1.448 cm2

Latihan Kendiri 1.7

1. (a) sSRT = 12π
= 37.70 m
SP + QT = 24 – 16
= 8 m
º Panjang pagar = 37.70 + 8 + 14 + 16
= 75.70 m

(b) sPR = 14

16q = 14

q = 0.875 rad

Luas sektor PQR = 1 (162)(0.875)
2

= 112 m2

Luas semibulatan SRT = 1 (122)(π)
2

= 226.22 m2

º Luas kawasan tanaman pokok bunga = 226.22 – 112

O = 114.22 m2
2. (a) OG = ! 122 – 92

= 7.937 cm 12 cm
º h = 12 – 7.937

= 4.063 cm

E 9 cm G

10

(b) sin ˙EOG = 9
12

˙EOG = 48° 35' 3.142
180°
˙EOF = 2(48° 35') ×

= 97° 10' × 3.142
180°
= 1.696 rad

Luas sektor EOF = 1 (122)(1.696)
2

= 122.11 cm2

Luas ∆EOF = 1 (12)(12)(sin 97° 10')
2

= 71.44 cm2

º Luas keratan rentas yang mengandungi air = 122.11 – 71.44
= 50.67 cm2

3. (a) kos ˙BAE = 4 A
18 4 cm

˙BAE = 77° 10' 18 cm

º ˙BAD = 77° 10'

(b) ˙BAD = 77° 10' × 3.142 E B
180°

= 1.347 rad

˙CBR = 2π – π – π – 1.347
2 2

= 1.795 rad

CD = BE = ! 182 – 42
= 17.55 cm

Luas trapezium ABCD = 1 (17.55)(7 + 11)
2

= 157.95 cm2

Luas sektor DAR = 1 (112)(1.347)
2

= 81.49 cm2

Luas sektor CBR = 1 (72)(1.795)
2

= 43.98 cm2

º Luas kawasan berlorek = 157.95 – 81.49 – 43.98
= 32.48 cm2

( ) 4. (a) q =20 3.142
60 × 360° × 180°

= 120° × 3.142
180°

= 2.095 rad

Luas sektor yang disurih = 1 (82)(2.095)
2

= 67.04 cm2
(b)
L = 80 cm2

1 (82)q = 80
2

32q = 80

q = 80
32

= 2.5 rad

11

Latihan Formatif 1.3

1. (a) sAB = 4.2 cm

6q = 4.2

q = 4.2
6

= 0.7 rad

(b) Luas sektor AOB = 1 (62)(0.7)
2

= 12.6 cm2

Luas sektor PAQ = 1 (32)(0.5)
2

= 2.25 cm2

º Luas kawasan berlorek = 12.6 – 2.25
= 10.35 cm2

2. (a) 60° = 60° × 3.142
180°

= 1.047 rad

(b) Luas sektor VOW = 1 (52)(1.047)
2

= 13.088 cm2

Luas ∆VOW = 1 (5)(5)(sin 60°)
2

= 10.825 cm2

º Luas tembereng berlorek VW = 13.088 – 10.825
= 2.263 cm2

3. (a) sPQ = 2πj
= 2(3.142)(3)

= 18.852 cm

OP = ! 32 + 42

=! 25

= 5 cm

5q = 18.852

q = 18.852
5

= 3.77 rad

(b) Luas sektor POQ = 1 (52)(3.77)
2

= 47.13 cm2

4. (a) sKL = 7 cm
4q = 7

q = 7
4

= 1.75 rad

(b) Sudut major KOL = 2π – 1.75

= 4.534 rad

º Luas sektor major KOL = 1 (42)(4.534)
2

= 36.27 cm2 A 9 cm O
70°
5. (a) AC = tan 70°
9
AC = 9(tan 70°)

= 24.73 cm

(b) Luas ∆AOC = 1 (24.73)(9)
2

= 111.285 cm2 C

12

º Luas OACB = 2(111.285)
= 222.57 cm2

(c) 140° = 140° × 3.142
180°

= 2.444 rad

º Luas sektor minor OAB = 1 (92)(2.444)
2

= 98.98 cm2

(d) Luas kawasan berlorek = 222.57 – 98.98

= 123.59 cm2

6. (a) RS = kos 60° SR
6 60°
RS = 6(kos 60°)

= 3 cm (tertunjuk)

(b) OR = ! 62 – 32 6 cm

=! 27

= 5.196 cm O

Luas ∆OSR = 1 (3)(5.196)
2

= 7.794 cm2

Luas OPSR = 2(7.794)

= 15.59 cm2 3.142
180°
Sudut major PSR = (360° – 120°) ×
=
240° × 3.142
180°

= 4.189 rad

Luas sektor major PQRS = 1 (32)(4.189)
2

= 18.85 cm2

º Luas panel OPQR = 15.59 + 18.85

= 34.44 cm2

(c) n = 5

Luas bulatan = 3.142(92)

= 254.502 cm2

Luas kawasan bukan panel = 254.502 – 5(34.44)

= 82.302 cm2

º Luas kawasan berlabel T = 82.302
5

= 16.46 cm2

Latihan Kendiri 1.8

1. (a) sin ˙AOP = 16
20

˙AOP = 53° 8' 3.142 O
180° P
º ˙AOB = 2(53° 8') ×

= 106° 16' × 3.142 20 cm
180° A 16 cm

= 1.855 rad

12q = 21

q = 21
12

= 1.75 rad

º ˙TSU = 1.75 rad

13

(b) sAQB = 20(1.855)
= 37.1 cm

sARB = 16(3.142)
= 50.27 cm

º Perimeter wau bulan = 37.1 + 50.27 + 21 + 2(12)

= 132.37 cm

(c) Luas sektor AOBQ = 1 (202)(1.855)
2

= 371 cm2

Luas ∆AOB = 1 × 32 × 12
2
= 192 cm2

Luas tembereng AQB = 371 – 192

= 179 cm2

Luas semibulatan APBR = 1 (162)(3.142)
2
= 402.18 cm2

Luas kawasan berlorek AQBR = 402.18 – 179

= 223.18 cm2

Luas sektor TSUR = 1 (122)(1.75)
2
= 126 cm2

º Luas wau bulan = 223.18 + 126
= 349.18 cm2

2. Katakan jejari duit syiling ialah j cm.

Luas ∆ABC = 1 (2 j)2(sin 60°) A
2 2j
=21 (4 j2)(0.866)

= 1.732 j2 60°
Bj D
60° = 60° × 3.142 C
180°

= 1.047 rad
[ ]Luas tiga sektor duit syiling = 3
1  j2(1.047)
2

= 1.571 j2

Luas kawasan berwarna biru = 12.842 mm2

1.732 j 2 – 1.571 j 2 = 12.842

0.161 j 2 = 12.842

j 2 = 12.842
0.161

= 79.764

j = ! 79.764

= 8.931 mm

Latihan Formatif 1.4

1. (a) (i) 40° = 40° × 3.142
180°

= 0.698 rad

sPQ = 11(0.698)
= 7.678 cm

º Perimeter sektor POQ = 2(11) + 7.678

= 29.68 cm

(ii) Luas sektor POQ = 1 (112)(0.698)
2

= 42.23 cm2

14

(iii) Isi padu sepotong kek = 8(42.23)

= 337.84 cm3

(b) Jisim sebiji kek = 9 × 150

= 1 350 gram

2. (a) tan ˙BEF = 6 F 6m B
4

˙BEF = 56° 19'

˙AEB = 2(56° 19') 4m
= 112° 38'

˙AED = ˙BEC = 360° – 2(112° 38') × 3.142 E
2 180°

= 67° 22' × 3.142
180°

= 1.176 rad

BE = ! 42­ + 62

=! 52

= 7.211 m

sAD = sBC = 7.211(1.176)
= 8.48 m

º Perimeter kolam renang = 2(12)+2(8.48)

= 40.96 m

(b) Luas segi empat tepat ABCD = 12(8)

= 96 m2

Luas tembereng AD = Luas sektor AED – Luas ∆AED

=21 (7.2112)(1.176) – 1 (7.211)(7.211)(sin 67° 22')
2

= 30.575 – 23.997

= 6.578 m2

º Luas lantai kolam renang = 96 + 2(6.578)

= 109.156 m2

(c) Isi padu air = 109.156(1.5)

= 163.734 m3

3. (a) kos ˙POS = 36 P S
46 36 cm

˙POS = 38° 30'

˙POQ = 2(38° 30') 46 cm
= 77°

q = 77° × 3.142 O
180°

= 1.344 rad

(b) sPRQ = 46(1.344)
= 61.824 cm

(c) Luas sektor POQ = 1 (462)(1.344)
2

= 1 421.952 cm2

Luas ∆POQ = 1 (46)(46)(sin 77°)
2

= 1 030.884 cm2

º Luas keratan rentas kayu = 1 421.952 – 1 030.884

= 391.068 cm2

15

4. (a) (i) 60° = 60° × 3.142
180°

= 1.047 rad

sAB = 30(1.047)
= 31.41 cm

(ii) Luas sektor COD = 1 (302)(1.047)
2

= 471.15 cm2

(iii) Perimeter tembereng EF = 31.41 + 30

= 61.41 cm

(iv) Luas ∆EOF = 1 (30)(30)(sin 60°)
2

= 389.71 cm2

Luas tembereng EF = 471.15 – 389.71

= 81.44 cm2

(b) Isi padu konkrit = 3(471.15)(5)

= 7 067.25 cm3

(c) Jumlah kos konkrit = 7 067.25(0.50)

= RM3 533.63

Latihan Sumatif

1. (a) Luas = 60 cm2

1 (102)q = 60
2
50q = 60

q = 60
50

= 1.2 rad

(b) sKL = 10(1.2)
= 12 cm

º Perimeter sektor KOL = 12 + 2(10)
= 32 cm

2. (a) 2 rad = 2 × 180°
3.142

= 114° 35'

CD2 = 32 + 32 – 2(3)(3)(kos 114° 35')
CD = 5.049 cm

sAB = 6(2)
= 12 cm

º Perimeter kawasan berlorek = 2(3) + 5.049 + 12

= 23.049 cm

(b) Luas sektor AOB = 1 (62)(2)
2
= 36 cm2

Luas ∆COD = 1 (3)(3)(sin 114° 35')
2

= 4.092 cm2

º Luas kawasan berlorek = 36 – 4.092

= 31.908 cm2

3. (a) Luas kawasan berlorek = 10.8 cm2

1 (62)q – 1 (42)q = 10.8
2 2

18q – 8q = 10.8

10q = 10.8

q = 10.8
10
= 1.08 rad
16

(b) sPQ = 4(1.08)
= 4.32 cm

OP = 2 = 4
PR 1 2

PR = 2 cm

OR = 2 + 4

= 6 cm

sRS = 6(1.08)
= 6.48 cm

º Perimeter kawasan berlorek = 2(2) + 4.32 + 6.48
= 14.8 cm

4. (a) Perimeter = 18 cm
2j + jq = 18

Luas = 8 cm2

1  j2q = 8
2

(b) 2j + jq = 18

j(2 + q) = 18

j= 18 …1
2+q
21  j2q = 8 …2

Gantikan 1 ke dalam 2:

1 18
2 2+q
( ) q  2=8

( ) 1 q  4 + 324 q 2 =8
2 4q +

162q = 8(4 + 4q + q 2)

162q = 32 + 32q + 8q 2

8q 2 – 130q + 32 = 0

4q 2 – 65q + 16 = 0

(4θ – 1)(q – 16) = 0
1
q = 4 atau q = 16 (abaikan)

Apabila q = 1 , j = 18 1
4 4
2 +

=(1948)

= 8

º j = 8 cm dan q = 1 rad
4

5. (a) sin ˙PCR = 4 P Q
5 5 cm

˙PCR = 53° 8' D 4 cm C

kos ˙DCQ = 4 4 cm 5 cm
5

˙DCQ = 36° 52'

º ˙PCQ = 53° 8' – 36° 52' RC

= 16° 16'

(b) DQ = PB = ! 52­ – 42

=! 9

= 3 cm

17

∠PCQ = 16° 16' × 3.142
180°

= 0.2839 rad

sPQ = 5(0.2839)
= 1.42 cm

º Perimeter kawasan berlorek APQ = 2(4 – 3) + 1.42

= 3.42 cm

(c) Luas segi empat sama ABCD = 4(4)
= 16 cm2

Luas ∆DCQ = Luas ∆CBP
=12 (3)(4)
= 6 cm2

Luas sektor PCQ = 1 (52)(0.2839)
2

= 3.55 cm2

º Luas kawasan berlorek APQ = 16 – 2(6) – 3.55
= 0.45 cm2

( ) 6. (a) sPR π
= 10 2

= 15.71 cm

sPQ = 2 × 15.71
5
10q = 6.284

q = 6.284
10

= 0.6284 rad

( )(b) ˙QOR = π – 180°
2 0.6284 × 3.142

= 0.9426 × 180°
3.142

= 54°

Luas sektor POQ = 1 (102)(0.6284)
2

= 31.42 cm2

Luas ∆QOR = 1 (10)(10)(sin 54°)
2

= 40.45 cm2

º Luas kawasan berwarna = 31.42 + 40.45
= 71.87 cm2

7. 60° = 60° × 3.142 Q
180°
= 1.047 rad 60°
1 P j cm
Luas sektor POQ = 2  j2(1.047)

= 0.524 j2

Luas ∆POQ = 1  j2 (sin 60°) O
2
= 0.433 j2

Luas tembereng PQ = 0.524 j2 – 0.433 j2
= 0.091 j2

120° = 120° × 3.142 R
180°
= 2.095 rad

120°
P j cm O

18

Luas sektor POR = 1  j2 (2.095)
2
= 1.048 j2
1
Luas ∆POR = 2  j2(sin 120°)

=12  j2(0.866)

= 0.433 j2

º Luas kawasan berlorek = 1.048 j2 – 0.433 j2 – 2(0.091 j2)

= 0.433 j2

8. sVW = 64(2)
= 128 cm

2πj = 128

j= 128
2π
=2(31.21842)
64 cm

= 20.369 cm

º Tinggi kon = ! 642­ – 20.3692 20.369 cm
= 60.67 cm

9. (a) AP = sin 30° P B
16
π–6 rad
AP = 16(sin 30°)

= 8 cm 16 cm

(b) BP = ! 162­ – 82

= 13.86 cm A

Luas ∆ABP = 1 (13.86)(8)
2

= 55.44 cm2 3.142
180°
(c) ˙AOP = (180° – 120°) × P
= 60° × 3.142
180° 30°
120°
= 1.047 rad O 8 cm B

Luas sektor AOP = 1 (82)(1.047)
2

= 33.504 cm2

Luas ∆ AOP = 1 (8)(8)(sin 60°)
2

= 27.713 cm2

º Luas kawasan berlorek = 33.504 – 27.713
= 5.791 cm2

10. (a) A(0, 8), B(6, 0), C(7, 7)

Luas ∆ ABC = 1   0 6 7 0
2 8 0 7 8

=12 (42 + 56 – 48)

= 25 unit2

(b) dAC = ! (7 – 8)2­ + (7 – 0)2
=! 50 unit

dBC = ! (7 – 0)2­ + (7 – 6)2
=! 50 unit

19

dAB = ! (0 – 8)­2 + (6 – 0)2 A Ίහ50 unit
=! 100 10 unit C
= 10 unit
Ίහ50 unit
102 = (! 50)2 + (! 50)2 – 2(! 50)(! 50)(kos ˙ACB)
(! 50)2 + (! 50)2 –
kos ˙ACB = 2(! 50)(! 50) 102

kos ˙ACB = 0

˙ACB = 90°
π
(c) ˙ACB = 2 rad B

( ) 1 (! 50 )2 π
Luas sektor ACB = 2 2

= 39.275 unit2

Luas ∆ ACB = 1 (! 50)(! 50)(sin 90°)
2

= 25 unit2

Luas tembereng AEB = 39.275 – 25

= 14.275 unit2

Luas semibulatan = 1 (52)(3.142)
2

= 39.275 unit2

º Luas kawasan berlorek = 39.275 – 14.275

= 25 unit2

11. (a) kos ˙DFH = 1 H
4
D
˙DFH = 75° 31'

˙BOD = 2(75° 31') × 3.142 1 unit 4 unit
180° F

= 151° 2' × 3.142
180°
= 2.636 rad

º q = 2.636 rad

(b) Luas sektor BFD = 1 (42)(2.636)
2

= 21.09 unit2

(c) Luas ∆ BFD = 1 (4)(4)(sin 151° 2')
2

= 3.874 unit2

º Luas kawasan berlorek = 21.09 – 2(3.874)

= 13.34 unit2
180°
12. (a) 1 rad = 1 × 3.142

= 57° 17'

JM 2 = 72 + 72 – 2(7)(7)(kos 57° 17')

JM = 6.711 cm

º Jejari bagi sektor bulatan JKLM ialah 6.711 cm.

(b) sJM = 7(1)
= 7 cm

sJKL = 6.711(3.8)
= 25.50 cm

º Perimeter rantau berlorek = 2(7) + 25.50

= 39.50 cm

(c) Luas sektor JAM = 1 (72)(1)
2

= 24.5 cm2

20

(d) Luas sektor JKLM = 1 (6.7112)(3.8)
2

= 85.571 cm2

Luas tembereng JM = 24.5 – 1 (7)(7)(sin 57° 17')
2

= 24.5 – 20.613

= 3.887 cm2

º Luas rantau berlorek = 85.571 – 2(3.887)

= 77.80 cm2

13. (a) 2 = sin 30° A
OP P 30°
2
OP = sin 30° 2 cm
O
= 4 cm

Jejari sektor PQR = 4 + 2

= 6 cm

sQR = jq
= 6(1.047)

= 6.282 cm

(b) AP = ! 42­ – 22

= 3.464 cm

Luas ∆APO = 1 (3.464)(2)
2

= 3.464 cm2 3.142
180°
Sudut sektor major AOB = (360° – 120°) ×
=
240° × 3.142
180°

= 4.189 rad

Luas sektor major AOB = 1 (22)(4.189)
2

= 8.378 cm2

Luas sektor PQR = 1 (62)(1.047)
2

= 18.846 cm2

º Luas kawasan berlorek = 18.846 – 2(3.464) – 8.378
= 3.54 cm2

14. (a) Luas sektor AOB = 243 m2
1
2 (182)q = 243

162q = 243

q = 243
162

= 1.5 rad

(b) 1.5 rad = 1.5 × 180°
3.142

= 85° 56'

AB2 = 182 + 182 – 2(18)(18)(kos 85° 56')
AB = 24.537 m

sACB = 12.269(π)
= 38.55 m

sAB = 18(1.5)
= 27 m

º Panjang pagar = 38.55 + 27
= 65.55 m

21

(c) Luas tembereng AB = 243 – 1 (18)(18)(sin 85° 56')
2
= 243 – 161.592

= 81.41 m2

Luas semibulatan ACB = 1 (12.2692)(π)
2
= 236.48 m2

º Luas kawasan pokok bunga = 236.48 – 81.41
= 155.07 m2

15. sAB = jq π A
2
= 5.5( )

= 8.641 cm

º Panjang tali = 4(11) + 4(8.641) O 5.5 cm B
= 44 + 34.564
= 78.564 cm

16. (a) π j = 110

j = 110
3.142

= 35 cm (tertunjuk)

(b) (i) 118° = 118° × 3.142
180°
= 2.06 rad

Luas sektor POQ = 1 (352)(2.06)
2
= 1 261.75 cm2

(ii) Luas ∆POQ = 1 (35)(35)(sin 118°)
2
= 540.805 cm2

º Luas tembereng berlorek = 1 261.75 – 540.805

= 720.945 cm2

(iii) Isi padu air = 720.945(200)

= 144 189 cm3

= 144.189 liter

17. (a) 40° = 40° × 3.142
180°
= 0.698 rad

sAB = 3(0.698)
= 2.094 cm

(b) Luas sektor AOB = 1 (32)(0.698)
2
= 3.141 cm2

(c) Isi padu prisma = 3.141(4)

= 12.564 cm3

(d) Luas satah ACEO = 3(4)

= 12 cm2

Luas permukaan melengkung ABDC = 2.094(4)
= 8.376 cm2

º Jumlah luas permukaan = 2(12) + 8.376 + 2(3.141)
= 38.658 cm2

22

18. (a) 60° = 60° × 3.142
180°

= 1.047 rad

Panjang setiap lengkok = 5(1.047)
= 5.235 cm

º Perimeter kawasan berwarna = 12(5.235)

= 62.82 cm
1
(b) Luas satu sektor = 2 (52)(1.047)

= 13.09 cm2

Luas satu segi tiga = 1 (5)(5)(sin 60°)
2
= 10.83 cm2

Luas satu tembereng = 13.09 – 10.83
= 2.26 cm2

º Luas kawasan berwarna = 12(2.26)
= 27.12 cm2

23

Jawapan

BAB 2 PEMBEZAAN

Aktiviti Penerokaan 1 (Halaman 30)

2. Nilai bagi f (0) tidak dapat ditentukan kerana pada x = 0, fungsi f (x) = x2 + 3x adalah tidak tertakrif.
x
3.
˜   Â

x – 0.1 – 0.01 – 0.001 – 0.0001 … 0.0001 0.001 0.01 0.1

f (x) 2.9 2.99 2.999 2.9999 … 3.0001 3.001 3.01 3.1

f (x)

6

4 f (x) = –x–2 –+x–3–x–
3

2

– 4 –2 0 x
24

Daripada jadual dan graf, nilai had bagi f (x) = x2 + 3x apabila x menghampiri sifar ialah 3, iaitu
x2 + 3x x
had f (x) = had x = 3.
x˜0 x˜0

4. had f (x) ≠ f (0)
x˜0

Latihan Kendiri 2.1

1. (a) had (x2 + x – 3) = 02 + 0 – 3
x ˜ 0 = –3

(b) had ! x + 1 = ! 0 + 1
x˜0
= ! 1


=1

( )(c) had x+4 = 0+ 4
x˜0 x–2 0– 2

= –2
( )(d) had
x˜0
a = a a
ax + a a(0) +
a
= a

= 1

2. (a) had (3x – 1) = 3(0) – 1
x ˜ 0 = –1

(b) had (! 10 – 2x ) = ! 10 – 2(–3)
x ˜ –3 = ! 16



=4

1

(c) had

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

= had (x – 2)
x ˜ –3

= –3 – 2

= –5
( ) ( )(d) had
x˜6
x–6 = had x–6
x2 – 36 x˜6 (x – 6)(x + 6)

( )= had 1
x˜6 x+6

= 6 1 6
+
1
= 12

(e) had

( ) ( )x ˜ 2
x2 – 3x + 2 = had (x – 2)(x – 1)
x2 – 4 (x + 2)(x – 2)
x˜2

( )= had x–1
x˜2 x+2

= 2 – 1
2 + 2
1
= 4

(f) had

( ) ( )( )x˜0
1 – ! 2x + 1 = had 1 – ! 2x + 1 1 + ! 2x + 1
2x2 – x x˜0 2x2 – x 1 + ! 2x + 1

( )( )= had
x˜0
1 – (2x + 1)
x(2x – 1) 1 + ! 2x + 1

( )( )= had
x˜0
–2x
x(2x – 1) 1 + ! 2x + 1

( )( )= had
x˜0
–2
(2x – 1) 1 + ! 2x + 1

= –2

(2(0) – 1)(1 + ! 2(0) + 1)
–2
= –2

= 1
(g) had
x–4 x–4 ! x + 2
( ) ( )( )x˜4! x – 2 ! x – 2 ! x + 2
= had
x˜4

( )= had
x˜4
(x – 4)(! x + 2)
x–4
= had (! x + 2)
x˜4

= ! 4 + 2
=2+2

=4

2

(h) had

( )( )x ˜ 3
3 – ! 2x + 3 = had 3 – ! 2x + 3 3 + ! 2x + 3
x–3 x˜3 x–3 3 + ! 2x + 3

9 – (2x + 3)
(x – 3) 3 + ! 2x + 3
( )= had
x˜3

6 – 2x
(x – 3) 3 + ! 2x + 3
( )=had

x˜3

–2(x – 3)
(x – 3) 3 + ! 2x + 3
( )= had
x˜3

= had –2
x˜3 3 + ! 2x + 3

= –2
3 + ! 2(3) + 3

= –2
3 + ! 9

= –2
6
= – 13

( )( )(i)
had x+2 –2 = had x+2 ! 5x + 14 + 2
! 5x + 14 x ˜ –2 ! 5x + 14 – 2 ! 5x + 14 + 2
x ˜ –2

= had (x + 2)(! 5x + 14 + 2)
(5x + 14) – 4
x ˜ –2

= had (x + 2)(! 5x + 14 + 2)
5x + 10
x ˜ –2

= had (x + 2)(! 5x + 14 + 2)
5(x + 2)
x ˜ –2

= had ! 5x + 14 +2
5
x ˜ –2

= ! 5(–2) + 14 +2
5

= ! 4 + 2
5

= 4
5

3. (a) had x2 – 2x = had x(x – 2)
x˜0 x 3 – 4x x˜0 x(x 2 – 4)
x(x – 2)
= had x(x + 2)(x – 2)

x˜0

= had x 1 2
+
x˜0

= 1
2
x2 – 4x + 3 (x – 3)(x – 1)
(b) had 2x 2 – 5x – 3 = had (2x + 1)(x – 3)
x˜3 x˜3

= had x–1
2x + 1
x˜3

= 3–1
2(3) + 1

= 2
7

3

(c) had x3 – 5x 2 + 6x = had x(x 2 – 5x + 6)
x˜3 x 2 – 3x x˜3 x(x – 3)

= had x(x – 2)(x – 3)
x˜3 x(x – 3)

= had x – 2
x˜3

= 3 – 2

= 1
(d) had

( )( )x˜0
3– 5x 9 = had 5x 3 + ! x + 9
! x + x˜0 3 – ! x + 9 3 + ! x + 9

= had 5x(3 + ! x + 9 )
x˜0 9 – (x + 9)

= had 5x(3 + ! x + 9 )
x˜0 – x

= had –5(3 + ! x + 9 )
x˜0

= –5(3 + ! 0 + 9)

= –5(6)

= –30
(e) had

( )( )x˜4
x–4 x = had x–4 2 + ! 8 – x
2 – ! 8 – x˜4 2 – ! 8 – x 2 + ! 8 – x

= had (x – 4)(2 + ! 8 – x)
4 – (8 – x)
x˜4

= had (x – 4)(2 + ! 8 – x)
x–4
x˜4

= had 2 + ! 8 – x
x˜4

= 2 + ! 8 – 4

= 2 + ! 4
=4

(f) had

( )( )x ˜ 7
! x + 2 – 3 = had ! x + 2 – 3 ! x + 2 + 3
x–7 x˜7 x–7 ! x + 2 + 3

(x + 2) – 9
(x – 7) ! x + 2 + 3
( )= had
x˜7

x–7
(x – 7) ! x + 2
( )= had +3
x˜7

= had 1 +3
x˜7 ! x + 2

= ! 7 1 + 3
+2

= 1
! 9 + 3

= 1
4. (a) (i) 4 6

(ii) Apabila x menghampiri sifar dari arah kiri, had f (x) = 1 dan apabila x menghampiri sifar dari arah kanan,
x˜0

had f (x) = 4. Oleh sebab had kiri tidak sama dengan had kanan, maka had f (x) tidak wujud.
x˜0 x˜0

(i) had f (x) = 2
x ˜ –1

(ii) had f (x) = 3
x˜5

4

Aktiviti Penerokaan 2 (Halaman 34)
5. Apabila titik C menghampiri titik B, nilai m menghampiri 6.
6.

dx x + dx y + dy dy dy
dx
1 4 16 7 7
3.25
0.5 3.5 12.25 0.3025 6.5
0.030025
0.05 3.05 9.3025 6.05

0.005 3.005 9.030025 6.005

7. Apabila dx menghampiri sifar, nilai dy menghampiri 6, iaitu:
dx
dy
Nilai had dx = Kecerunan garis tangen di titik B

dx ˜ 0
=6

Latihan Kendiri 2.2

1. (a) y = x

dy = x + dx – x

= dx

dy = 1
dx

Maka, ddxy = had dy
dx
dx ˜ 0

= 1

(b) y = 5x

dy = 5(x + dx) – 5x

= 5x + 5dx – 5x

dy = 5dx
dx = 5

Maka, ddyx = had dy
dx
dx ˜ 0

= 5

(c) y = – 4x

dy = – 4(x + dx) – (– 4x)

= – 4x – 4dx + 4x

= – 4dx

dy = – 4
dx
Maka, ddxy = dy
had dx

dx ˜ 0

= – 4

(d) y = 6x2

dy = 6(x + dx)2 – 6x2

= 6(x2 + 2xdx + (dx)2) – 6x2

= 6x2 + 12xdx + 6(dx)2 – 6x2

= 12xdx + 6(dx)2

dy = 12x + 6dx
dx
Maka, ddxy =
had (12x + 6dx)

dx ˜ 0

= 12x

5

(e) y = –x2

dy = –(x + dx)2 – (–x2)

= –(x2 + 2xdx + (dx)2) + x2

= –x2 – 2xdx – (dx)2 + x2

= –2xdx – (dx)2

dy = –2x – dx
dx
Maka, ddxy =
had (–2x – dx)

dx ˜ 0

= –2x

(f) y = 2x3

dy = 2(x + dx)3 – 2x3

= 2(x + dx)(x + dx)2 – 2x3

= 2(x + dx)(x2 + 2xdx + (dx)2) – 2x3

= 2[x3 + 2x2dx + x(dx)2 + x2dx + 2x(dx)2 + (dx)3] – 2x3

= 2[x3 + 3x2dx + 3x(dx)2 + (dx)3] – 2x3

= 2x3 + 6x2dx + 6x(dx)2 + 2(dx)3 – 2x3

= 6x2dx + 6x(dx)2 + 2(dx)3

dy = 6x2 + 6xdx + 2(dx)2
dx
Maka, ddyx = dy
had dx

dx ˜ 0

= had (6x2 + 6xdx + 2(dx2))
dx ˜ 0

1 = 6x2
2
(g) y = x2 = 1 (x + dx)2 – 1 x2
dy 2 2

= 1 (x2 + 2xdx + (dx)2) – 1 x2
2 2
1 1 1
= 2 x2 + xdx + 2 (dx)2 – 2 x2

= xdx + 1 (dx)2
= x+ 2
dy 1
dx 2 dx

Maka, ddxy = had dy
dx
dx ˜ 0

= had (x + 1 dx)
2
dx ˜ 0

1 =x
x
(h) y = dy = 1 – 1
x + dx x

= x – (x + dx)
x(x + dx)

= x – x – dx
x(x + dx)

= –dx
x(x + dx)
dy
dx = – x(x 1 dx)
+

Maka, ddyx = had dy
dx
dx ˜ 0
[ ]= had
dx ˜ 0 – x(x 1 dx)
+
1
= –  x2

6

2. y = 2x2 – x + 7

dy = 2(x + dx)2 – (x + dx) + 7 – (2x2 – x + 7)

= 2(x2 + 2xdx + (dx)2) – x – dx + 7 – 2x2 + x – 7

= 2x2 + 4xdx + 2(dx)2 – x – dx + 7 – 2x2 + x – 7

= 4xdx + 2(dx)2 – dx

dy = 4x + 2dx – 1
dx

Maka, dy = had dy
dx dx
dx ˜ 0

= had (4x + 2dx – 1)
dx ˜ 0

= 4x – 1

3. y = 3 + x – x2

dy = 3 + x + dx – (x + dx)2 – (3 + x – x2)

= 3 + x + dx – (x2 + 2xdx + (dx)2) – 3 – x + x2

= dx – 2xdx – (dx)2

dy = 1 – 2x – dx
dx

Maka, dy = had dy
dx dx
dx ˜ 0

= had (1 – 2x – dx)
dx ˜ 0

= 1 – 2x

Latihan Formatif 2.1

1. (a) (i) 8 (ii) 3

(iii) 0 (iv) –1

(v) 0 (vi) had x2 – 4x + 3 = 42 – 4(4) + 3
x ˜ 4
=3

(b) had x2 – 4x + 3 = 8
x ˜ a a2 – 4a + 3 = 8

a2 – 4a – 5 = 0

(a + 1)(a – 5) = 0

a = –1 atau a = 5

(c) (i) dy = 2x – 4
dx

(ii) 2(4) – 4 = 4

2. (a) had (x2 – 6x + 9) = 02 – 6(0) + 9
x ˜ 0
=9

(b) had 3! x 4 – 2x2 = 3! 24 – 2(2)2
x˜2
= 3! 8


=2
[ ] [ ](c) had
x˜9
9–x = had –(x – 9)
x 2 – 81 x˜9 (x – 9)(x + 9)

= – 9 1 9
+

= – 118

[ ] [ ](d) had
x˜2
x2 – x – 2 = had (x – 2)(x + 1)
x–2 x˜2 x–2

= had (x + 1)
x˜2

=2+1
=3

7

[ ] [ ](e) had
x˜1
x3 – x = had x(x2 – 1)
x–1 x˜1 x–1

[ ]= had
x˜1
x(x – 1)(x + 1)
x–1

= had x(x + 1)
x˜1

= 1(1 + 1)

= 2

(f) had

[ ] [ ]x ˜ 5
x2 – 7x + 10 = had (x – 2)(x – 5)
x2 – 25 x˜5 (x – 5)(x + 5)

[ ]= had x–2
x˜5 x+5

= 5–2
5+5
3
= 10

3. (a) had

( )( )x ˜ 0
! 1 + 2x – ! 1 – 2x = had ! 1 + 2x – ! 1 – 2x ! 1 + 2x + ! 1 – 2x
x x˜0 x ! 1 + 2x + ! 1 – 2x x

1 + 2x – (1 – 2x)
x ! 1 + 2x + ! 1 – 2x
( )= had
x˜0

4x
x ! 1 + 2x + ! 1 – 2x
( )= had
x˜0

= had ! 1 + 2x 4
x˜0 + ! 1 – 2x

= 4
! 1 + 2(0) + ! 1 – 2(0)

= ! 1 4
+ ! 1

= 4
2

= 2

(b) had

( )( )x ˜ 4
3 – ! x + 5 = had 3 – ! x + 5 3 + ! x + 5
x–4 x˜4 x–4 3 + ! x + 5

9 – (x + 5)
(x – 4) 3 + ! x + 5
( )= had
x˜4

4–x
(x – 4) 3 + ! x + 5
( )= had
x˜4

–(x – 4)
(x – 4) 3 + ! x + 5
( )= had
x˜4

= had –1
x˜4 3 + ! x + 5

= –1
3 + ! 4 + 5

= –1
3 + ! 9

= – 16

8

(c) had

( )( )x˜3
x 2 – 5x + 6 = had x 2 – 5x + 6 2 + ! x + 1
2 – ! x + 1 2 – ! x + 1 2 + ! x + 1
x˜3

( )=
had (x 2 – 5x + 6) 2 + ! x + 1
4 – (x + 1)
x˜3

( )= (x – 2)(x – 3) 2 + ! x + 1
had 3– x

x˜3

( )= ! x
had (x – 2)(x – 3) 2 + + 1
–(x – 3)
x˜3

= had –(x – 2)(2 + ! x + 1)
x˜3

= –(3 – 2)(2 + ! 3 + 1)
= –(2 + ! 4 )
= – 4

4. (a) Secara penggantian langsung menghasilkan bentuk tak tentu 0 .
Jadi, 22 – k = 0 0

4 – k = 0

k = 4

(b) (−1)2 − 2(−1) − h = 0

3 − h = 0

h=3

k(−1) + 2 = 0

−k + 2 = 0

k=2

Jadi, h + k = 3 + 2

= 5

5. (a) y = 5x – 8

dy = 5(x + dx) – 8 – (5x – 8)

= 5x + 5dx – 8 – 5x + 8

= 5dx

dy = 5
dx
Maka, ddyx = dy
had dx

dx ˜ 0

= had 5
dx ˜ 0

= 5

(b) y = x2 – x

dy = (x + dx)2 – (x + dx) – (x2 – x)

= x2 + 2xdx + (dx)2 – x – dx – x2 + x

= 2xdx + (dx)2 – dx

dy = 2x + dx – 1
dx
Maka, ddyx = dy
had dx

dx ˜ 0

= had (2x + dx – 1)
dx ˜ 0

= 2x – 1

(c) y = (x + 1)2

= x2 + 2x + 1

dy = (x + dx)2 + 2(x + dx) + 1 – (x2 + 2x + 1)

= x2 + 2xdx + (dx)2 + 2x + 2dx + 1 – x2 – 2x – 1

= 2xdx + (dx)2 + 2dx

9

dy = 2x + dx + 2
dx
Maka, ddyx = dy
had dx

dx ˜ 0

= had (2x + dx + 2)
dx ˜ 0

= 2x + 2
1
(d) y = 4x dy = 1 – 1
4(x + dx) 4x

= 4x – 4x – 4dx
16x(x + dx)

dy = – 16x(4xd+x dx)
dx = 1
– 4x(x + dx)

Maka, ddyx = had dy
dx
dx ˜ 0

[ ]= had 1
dx ˜ 0 +
– 4x(x dx)

= – 41x2

6. d s = (t + d t)2 – 3(t + d t) – (t 2 – 3t)

= t 2 + 2td t + (d t)2 – 3t – 3d t – t 2 + 3t

= 2td t + (d t)2 – 3d t
ds
d t = 2t + d t – 3

dd st = had ds
d t
dt ˜ 0

= had (2t + d t – 3)
dt ˜ 0

= 2t – 3 ds
d t
Apabila t = 5, = 2(5) − 3

=7

Maka, halaju tupai pada t = 5 ialah 7 ms−1.

Aktiviti Penerokaan 3 (Halaman 39)

6. Bentuk graf bagi y = f (x) ialah parabola manakala bentuk graf bagi fungsi kecerunannya y = f (x) merupakan
suatu garis lurus.
Apabila nilai a berubah, garis tangen pada graf y = f (x) = x2 juga turut berubah.

7.

Koordinat-x –3 –2 –1 0 1 2 3

Kecerunan lengkung –6 – 4 –2 0 2 4 6

8. f(x) = x2
f (x) = 2x2 – 1

= 2x

f (–3) = 2(–3) f (–2) = 2(–2) f (–1) = 2(–1)
= –6 = – 4 = –2

f (0) = 2(0) f (1) = 2(1) f (2) = 2(2) f (3) = 2(3)
=0 =2 =4 =6

9. Bentuk graf bagi fungsi kubik y = f (x) = x3 ialah lengkung kubik manakala bentuk graf bagi fungsi kecerunannya
y = f (x) = 3x2 merupakan parabola.

10

Latihan Kendiri 2.3
( ) 1.
(a) d 4 x10 = 4 (10x10 – 1)
dx 5 5
4
= 5 (10x9)

= 8x9
ddx(–2x 4) ==
(b) –2(4x 4 – 1)
–2(4x3)

= –8x3

( ) ( )(c) d 3 = d 3 x–8
dx 4x8 dx 4

= 3 (–8x–8 – 1)
4

= 3 (–8x–9)
4
= – x69

( ) ( ( ) )(d) d6 = d 6x– 31
dx 3! x dx
6 – 13 x– 31
= – 1

= –2x– 43

= –  2
4
x 3
= – 3!2 x4

( )( )(e)
d – 123! x2 = d 2
dx dx
–12x3

( ) = –12 2 2 – 1
3
x3

= –8x– 13

= – 3!8 x

2. (a) d (4x2 + 6x – 1) = d (4x2) + d (6x) – d (1) Â 1 = x 0
dx dx dx dx

= 4(2x2 – 1) + 6(1x1 – 1) – 0x0 – 1

= 8x + 6

( ) ( ) ( )(b) d4 ! x + 2 = d 4 1 + d 2x– 21
dx 5 ! x dx 5 dx
x 2

( ) ( ) 1
= 4 1 – 1 + 2 – 12 x– 12 – 1
5 2 x2

= 2 x– 12 – x– 32
5

= 2 – 1
5! x ! x3

(c) d (9 – 4x)2 = d (81 – 72x + 16x2)
dx dx

= d (81) – d (72x) + d (16x2)
dx dx dx

= 32x – 72

11

( )( ) 3. (a) d ! x d 5
dx 4x2(5 – = dx 20x2 –
4x2

( ) = 20(2x2 – 1) – 4 5 5 – 1
2
x2

3
= 40x – 10x2

= 40x – 10! x3

( )(b) d x2 + 4 2= d (x 4  + 8x + 16x –2)
dx x dx

= 4x 4 – 1  + 8(1x1 – 1) + 16(–2x –2 – 1)

= 4x3 + 8 – 32
x3
( ) ( )(c)
d (4x – 1)(1 – x) = d 4x – 4x2 – 1 + x
dx ! x dx 1
x2

( ) = d 1 3 – x – 12 + 1
dx
4x2  – 4x2 x 2

( ) = d 1 3 – x – 21
dx
5x2  – 4x2

( ) ( ) ( )
=5 1 1 – 1 –4 3 3 – 1 – – 21 x– 21  – 1
2 2
x2  x2 

= 5 – 6! x + 1
2! x 2! x3
4. (a) y = x2 – 2x

dy = 2x – 2
dx
( )Apabila x = 1 dy 1
2 , dx =2 2 –2

= –1
(b) y = ! x (2 – x)

13

= 2x2 – x2

dy = 1 – 3 ! x
dx ! x 2
dy
Apabila x = 9, dx = 1 – 3 ! 9
! 9 2

= 1 – 9
3 2

= – 265

x2 + 4 = – 4  1
x2 6
(c) y =

= 1 + 4x–2

dy = –8x –3
dx
= – x83
dy
Apabila x = 2, dx = – 283

= – 8
8

= –1

12

Aktiviti Penerokaan 4 (Halaman 42)

2. y = (2x + 3)2

= (2x + 3)(2x + 3)

= 4x2 + 12x + 9
dy
dx = 8x + 12

3. (a) y = u2 du dy
dx du
(b) u = 2x + 3, jadi = 2 dan y = u2. Maka, = 2u

(c) dduy × du = 2u × 2
dx = 4u

= 4(2x + 3)

= 8x + 12

4. Jawapannya adalah sama.
Kaedah dalam langkah 3 menjadi pilihan kerana lebih mudah dan cepat untuk memperoleh terbitannya.

Latihan Kendiri 2.4

1. (a) d (x + 4)5 = 5(x + 4)5 – 1 d (x + 4)
dx dx

= 5(x + 4)4 d
d dx
(b) dx (2x – 3)4 = 4(2x – 3)4 – 1 (2x – 3)

= 4(2x – 3)3(2)

= 8(2x – 3)3

[ ](c) d 1 (6 – 3x)6 = 1 [6(6 – 3x)6 – 1] d (6 – 3x)
dx 3 3 dx

= 2(6 – 3x)5(–3)

= –6(6 – 3x)5

(d) d (4x2 – 5)7 = 7(4x2 – 5)7 – 1 d (4x2 – 5)
dx dx

= 7(4x2 – 5)6(8x)

= 56x(4x2 – 5)6

( ) ( ) ( )(e) d 1 x + 2 8 = 8 1 x + 2 8 – 1d 1 x + 2
dx 6 6 dx 6

( ) ( ) = 8 1 x + 2 7 1
6 6

( ) = 4 1 x + 2 7
3 6

[ ](f) d 2 (5 – 2x)9 = 2 [9(5 – 2x)9 – 1] d (5 – 2x)
dx 3 3 dx

= 6(5 – 2x)8(–2)

= –12(5 – 2x)8

(g) d (1 – x – x2)3 = 3(1 – x – x2)3 – 1 d (1 – x – x2)
dx dx

= 3(1 – x – x2)2(–2x – 1)

= –3(2x + 1)(1 – x – x2)2

(h) d (2x3 – 4x + 1)–10 = –10(2x3 – 4x + 1)–10 – 1 d (2x3 – 4x + 1)
dx dx

= –10(2x3 – 4x + 1)–11(6x2 – 4)

= –  60x2 – 40
(2x3 – 4x + 1)11

= –  20(3x2 – 2)
(2x3 – 4x + 1)11

13

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

= –1(3x + 2)–2(3)

= – (3x 3 2)2
+
( )(b)
d 1 = d [(2x – 7)–3]
dx (2x – 7)3 dx
d
= –3(2x – 7)–3 – 1 dx (2x – 7)

= –3(2x – 7)–4(2)

= – (2x 6 7)4
–

 ( )(c)
d 5 = ddx [5(3 – 4x)–5] d
dx (3 – 4x)5 dx

= 5[–5(3 – 4x)–5 – 1] (3 – 4x)]

= –25(3 – 4x)–6(– 4)

= 100
(3 – 4x)6

( ) [ ](d) d 3
dx 3 = d 4 (5x – 6)–8
4(5x – 6)8 dx

[ ] = 3 d
4 –8(5x – 6)–8 – 1 dx (5x – 6)

= –6(5x – 6)–9(5)

= – (5x3–0 6)9

(e) d (! 2x – 7) = d (2x – 1
dx dx
7)2

= 1 (2x – 1 – 1 d (2x – 7)
2 dx
7)2

= 1 (2x – 7)– 21(2)
2
1
= ! 2x – 7

(f) d (! 6 – 3x ) = d (6 – 1
dx dx
3x)2

= 1 (6 – 1 – 1 d (6 – 3x)
2 dx
3x)2

= 1 (6 – 3x)– 12(–3)
2
3
= – 2! 6 – 3x

( )(g) d d 1
dx ! 3x2 + 5 = dx (3x2 +
5)2

= 1 (3x2 + 1 – 1 d (3x2 + 5)
2 dx
5)2

= 1 (3x2 + 5)– 12(6x)
2
3x
= ! 3x2 + 5

14

( )(h) d ! x2 – x + 1 = d (x2 – x + 1
dx dx
1)2

= 1 (x2 – x + 1 – 1 d (x2 – x + 1)
2 dx
1)2

= 1 (x2 – x + 1)– 12(2x – 1)
2
2x – 1
= 2! x2 – x + 1

3. (a) y = (2x + 5)4

dy = 4(2x + 5)4 – 1 d (2x + 5)
dx dx

= 4(2x + 5)3(2)

= 8(2x + 5)3

Apabila x = 1, dy = 8[2(1) + 5]3
dx
= 8(343)

= 2 744
(b) y = ! 5 – 2x

1

y = (5 – 2x)2

dy = 1 (5 – 1 – 1 d (5 – 2x)
dx 2 dx
2x)2

= 1 (5 – 2x)– 12(–2)
2
1
= –  ! 5 – 2x

Apabila x = 1 , dy = –  1
2 dx ! 4

1 = – 12
2x –
(c) y = 3

= (2x – 3)–1

dy = –1(2x – 3)–1 – 1 d (2x – 3)
dx dx

= –1(2x – 3)–2(2)

= – (2x 2 3)2
–

Apabila y = 1, 1 = 1 3
2x –

2x – 3 = 1

2x = 4

x=2

dan dy = – [2(2)2– 3]2
dx

= –2

Aktiviti Penerokaan 5 (Halaman 44)

2. y = (x2 + 1)(x – 4)2

= (x2 + 1)(x – 4)(x – 4)

= (x2 + 1)(x2 – 8x + 16)

= x 4 – 8x3 + 16x2 + x2 – 8x + 16

= x 4 – 8x3 + 17x2 – 8x + 16

dy = 4x3 – 24x2 + 34x – 8
dx

15

3. (a) u = x2 + 1, jadi du = 2x dan v = (x – 4)2. Maka, dv = 2(x – 4).
dx dx
(b) u ddxv + v ddux = (x2 + 1) × 2(x – 4) + (x – 4)2 × 2x
= 2(x2 + 1)(x – 4) + 2x(x – 4)2

= (x – 4)[2(x2 + 1) + 2x(x – 4)]

= (x – 4)(2x2 + 2 + 2x2 – 8x)

= (x – 4)(4x2 – 8x + 2)

= 4x3 – 8x2 + 2x – 16x2 + 32x – 8

= 4x3 – 24x2 + 34x – 8

4. Jawapannya adalah sama.
Kaedah dalam langkah 3 menjadi pilihan kerana tidak perlu mengembangkan dua ungkapan algebra yang diberi
terlebih dahulu.

Aktiviti Penerokaan 6 (Halaman 45)

2. y = x(x – 1)–2 d u ddxv v ddux
dx
Menggunakan (uv) = +

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

= –2x + x – 1
(x – 1)3

= –x – 1
(x – 1)3

= – (xx +1
– 1)3

3. (a) u = x, jadi du = 1 dan v = (x – 1)2. Maka, dv = 2(x – 1) = 2x – 2.
dx dx
v ddux – u ddvx
(b) v2 = (x – 1)2(1) – x(2x – 2)
(x – 1)4

= x2 – 2x + 1 – 2x2 + 2x
(x – 1)4

= 1 – x2
(x – 1)4

= (1 + x)(1 – x)
(x – 1)4

= –(x – 1)(1 + x)
(x – 1)4

= – (xx + 1
– 1)3

4. Jawapannya adalah sama. menjadi pilihan kerana boleh mencari dy secara langsung tanpa menggunakan petua
Kaedah dalam langkah 3 dx
hasil darab.

Perbincangan (Halaman 45) u
v
Pertimbangkan hasil bahagi dua polinomial y = , dengan u = f (x) dan v = g(x).

y = u …1
v
u + du
Daripada prinsip pertama: y + dy = v + dv …2

2 – 1: dy = u + du – u
v + dv v

= v(u + du) – u(v + dv)
v(v + dv)

=
vdu – udv …3
v(v + dv)

16

B ahagikan 3 dengan dx: ddxy = v vdd(uxv +– ud vdd)xv
Ambil had di ked ua-dua belah: dhx a˜d0 ddxy = vdhx a˜ vdd0hx dda˜dux0(–v u+dhxda˜vd)0 ddvx

Oleh sebab dx ˜ 0, maka dv juga mendekati sifar, jadi had (v + dv) = v.
dx ˜ 0

had dy = vdhx a˜d0 ddux – udhx a˜d0 ddvx
dx v2
dx ˜ 0



Oleh sebab had dy = dy , maka had du = du dan had dv = dv .
dx dx dx dx dx dx
dx ˜ 0 dx ˜ 0 dx ˜ 0

( )Maka, petua hasil bahagi menggunakan idea had ialah: d u = v ddux – u ddvx
dx v v2

Perbincangan (Halaman 46)

1. Kaedah 1:

y = x(1 – x2)2

= x(1 – 2x2 + x4)

= x – 2x3 + x5
dy
dx = 1 – 6x2 + 5x4

Kaedah 2:

y = x(1 – x2)2 d u ddxv v ddux
dx
Menggunakan (uv) = +

dy = x(2)(1 – x2)(–2x) + (1 – x2)2(1)
dx
= – 4x2(1 – x2) + 1 – 2x2 + x4

= – 4x2 + 4x4 + 1 – 2x2 + x4

= 1 – 6x2 + 5x4

Ya, jawapannya adalah sama.

2. (a) Petua rantai:
y = 3(2x – 1)4

dy = 3(4)(2x – 1)4 – 1 ddx (2x – 1)
dx
= 12(2x – 1)3(2)

= 24(2x – 1)3

(b) Petua hasil darab:
d u ddvx v ddux
dx (uv) = +

dy = 3(4)(2x – 1)3(2) + (2x – 1)4(0)
dx
= 24(2x – 1)3

º Petua rantai

Latihan Kendiri 2.5

1. (a) y = 4x2(5x + 3)

dy = u dv + v du
dx dx dx

= 4x2(5) + (5x + 3)(8x)

= 20x2 + 40x2 + 24x

= 60x2 + 24x

17

(b) y = –2x3(x + 1)

dy = u dv + v du
dx dx dx

= –2x3(1) + (x + 1)(–6x2)

= –2x3 – 6x3 – 6x2

= –8x3 – 6x2

(c) y = x2(1 – 4x)4

dy = u dv + v du
dx dx dx

= x2[4(1 – 4x)3(– 4)] + (1 – 4x)4(2x)

= –16x2(1 – 4x)3 + 2x(1 – 4x)4

= 2x(1 – 4x)3[–8x + (1 – 4x)]

= 2x(1 – 12x)(1 – 4x)3

(d) y = x2! 1 – 2x2

dy = u dv + v du
dx dx dx
2x2)– 12(– 4x)
[ ] ( ) = x21
2 (1 – + ! 1 – 2x2 (2x)

= –  ! 1 2x3 + 2x! 1 – 2x2
– 2x2

= –2x3 + 2x(1 – 2x2)
! 1 – 2x2

= –2x3 + 2x – 4x3
! 1 – 2x2

= –6x3 + 2x
! 1 – 2x2

= 2x(1 – 3x2)
! 1 – 2x2

(e) y = (4x – 3)(2x + 7)6

dy = u dv + v du
dx dx dx

= (4x – 3)[6(2x + 7)5(2)] + (2x + 7)6(4)

= 12(4x – 3)(2x + 7)5 + 4(2x + 7)6

= 4(2x + 7)5[3(4x – 3) + (2x + 7)]

= 4(2x + 7)5[12x – 9 + 2x + 7]

= 4(14x – 2)(2x + 7)5

= 8(7x – 1)(2x + 7)5

(f) y = (x + 5)3(x – 4)4

dy = u dv + v du
dx dx dx

= (x + 5)3[4(x – 4)3(1)] + (x – 4)4[3(x + 5)2(1)]

= 4(x + 5)3(x – 4)3 + 3(x – 4)4(x + 5)2

= (x + 5)2(x – 4)3[(4(x + 5) + 3(x – 4)]

= (x + 5)2(x – 4)3(4x + 20 + 3x – 12)

= (7x + 8)(x + 5)2(x – 4)3

2. (a) y = (1 – x2)(6x + 1)

dy = u dv + v du
dx dx dx

= (1 – x2)(6) + (6x + 1)(–2x)

= 6 – 6x2 – 12x2 – 2x

= 6 – 18x2 – 2x

= –2(9x2 + x – 3)

18

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

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

(c) y = (x3 – 5)(x2 – 2x + 8)
= x5 – 2x4 + 8x3 – 5x2 + 10x – 40

dy = 5x4 – 8x3 + 24x2 – 10x + 10
dx
3. f (x) = x! x – 1
1)– 12(1)
[( ) ] ( )f (x) = x1
2 (x – + ! x – 1 (1)

= x + ! x – 1
2! x – 1

= x + 2(x – 1)
2! x – 1

= 3x – 2
2! x – 1

f (5) = 3(5) – 2
2! 5 – 1

= 13
4

4. y = x! x2 + 9 9)– 12(2x)
[ ] ( )dy
1
dx 2
= x (x2 + + ! x2 + 9 (1)

= x2 + ! x2 + 9
! x2 + 9

= x2 + x2 + 9
! x2 + 9

= 2x2 + 9
! x2 + 9

Pada x = 4, dy = 2(4)2 + 9
dx ! 42 + 9

= 41
5

( ) 5. (2x – 7)(0) – 3(2)
(a) d 3 = (2x – 7)2
dx 2x – 7

= – (2x 6 7)2
–

( )(b) (4x + 6)(3) – 3x(4)
d 3x = (4x + 6)2
dx 4x + 6

= 12x + 18 – 12x
(4x + 6)2

= 18
(4x + 6)2

( )(c) (1 – 6x)(8x) – (4x2)(–6)
d 4x2 = (1 – 6x)2
dx 1 – 6x

= 8x – 48x2 + 24x2
(1 – 6x)2
8x – 24x2
= (1 – 6x)2

= 8x(1 – 3x)
(1 – 6x)2

19

( )(d) (2x – 1)(3x2) – (x3 + 1)(2)
d x3 + 1 = (2x – 1)2
dx 2x – 1

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

( ) ( )(e) d! x (x + 1) 1 – (! x )(1)
dx x+1 2! x
= (x + 1)2

( )x + 1 – 2x
= 2! x
(x + 1)2

= 1–x 1)2
2! x (x +
( ) ( )(f)
d x (! x – 1)(1) – 1 (x)
dx ! x – 1 2! x – 1
= (! x – 1 )2

( ) =
2(x – 1) – x

2! x – 1
x–1
x–2
= 2(x – 1)! x – 1

= x–2
2! (x – 1)3

( ) ( )(g) ddx 2x
( ) ( )= ! 2x2 + 3
3x2 ! 2x2 + 3 (6x) – (3x2)
! 2x2 + 3
! 2x2 + 3 2

( )6x(2x2 + 3) – 6x3
= ! 2x2 + 3
2x2 + 3

= 12x3 + 18x – 6x3
! 2x2 + 3 (2x2 + 3)

= 6x3 + 18x
! 2x2 + 3 (2x2 + 3)

= 6x(x2 + 3)
! (2x2 + 3)3

(   ) ( )!  (h)
d 4x + 1 = d ! 4x + 1
dx 3x2 – 7 dx ! 3x2 –7

( ) ( ) 2
(! 3x2 – 7 ) ! 4x + 1 – (! 4x + 1) 3x
! 3x2 – 7
=
(! 3x2 – 7 )2

( )2(3x2 – 7) – 3x(4x + 1)
= (! 4x + 1 )(! 3x2 – 7 )

3x2 – 7

= 6x2 – 14 – 12x2 – 3x

(3x2 – 7)(! 4x + 1 )(! 3x2 – 7 )
6x2 – 14 – 12x2 – 3x
= ! 4x + 1 ! (3x2 – 7)3
( )
6x2 + 3x + 14
! 4x + 1 ! (3x2 – 7)3
( ) – 
=

20

6. 2(x + 5) – (2x – 3) = 2x + 10 – 2x + 3
(x + 5)2 (x + 5)2

= 13
(x + 5)2

Jadi, secara perbandingan, nilai pemalar r ialah 13.

Latihan Formatif 2.2

( ) 1. (a) d 9x2 – 3 = d (9x2 – 3x–2)
dx x2 dx

= 18x + 6x–3

= 18x + 6
x3

( )(b) d 6 – 1 + 8 = d (6x–3 – x–1 + 8)
dx x3 x dx

= –18x– 4 + x–2

= 1 – 18
x2 x4
1
[ ](c) d 5x + 4! x –7 = d (5x + – 7)
dx dx 4x2

= 5 + 2x– 21
2
= 5 + ! x

( ) ( )(d) d10 + 3 = d 10x– 21 + 3x– 31
dx ! x 3! x dx

= –5x– 32 – x– 43

= –  5 – 1
3 4
x2 x3

= – !5 x3 – 1
3! x 4

( ) ( )(e) dx2– 3 2= d x4 – 6x + 9
dx x dx x2
d
= dx (x4 – 6x + 9x –2)

= 4x3 – 6 – 18
x3
( ) ( )(f) 3 1
d 8x2 + x = d +
dx ! x dx 8x2 1 x2
2 x– 12
= 1

12x2 +

= 12! x + 1
2! x

( ) ( )(g) d4 – πx + 6 = d 4 x –3 – πx + 6
dx 9x3 dx 9
= – 43 x – 4 – π
= – 34x4 – π

[ ](h) d d 1 3
dx ! x (2 – x) = dx (2x 2 –
x2)

= x– 12 – 3 1
2
x2

= 1 – 3 ! x
! x 2

21

2. f (x) = 2 + 6x– 13

3x3

f (x) = 2x– 13 – 2x– 43

= 2 – 2
3! x 3! x 4

f (8) = 2 – 2
3! 8 3! 84
2
= 1 – 16

= 7
8
3. (a) f (t) = 6t3
3! t
1
= 6t 3 – 3

8
= 6t3
( )(b) 8 8 – 1
f (t) = 6 3
t3

5
= 16t3

( )(c) f ( 1 ) = 16 1 5
8 8
3

( ) = 16 1 5
23 3

= 24
25

= 1
2

4. s = 3t2 + 5t – 7

ds = 6t + 5
dt
ds
Apabila dt , 0

6t + 5 , 0

6t , –5
t , – 65
5. y = ax3 + bx2 + 3

Pada titik (1, 4), 4 = a(1)3 + b(1)2 + 3

4=a+b+3
a + b = 1 …1

dan dy = 7
dx

3ax2 + 2bx = 7

Pada titik (1, 4), 3a(1)2 + 2b(1) = 7
3a + 2b = 7 …2
2 × 1: 2a + 2b = 2 …3
2 – 3:
a=5

Gantikan a = 5 dalam 1: 5 + b = 1

b = – 4
º a = 5 dan b = – 4

22

6. dy = 3
dx

3x2 – 6x + 6 = 3
3x2 – 6x + 3 = 0
x2 – 2x + 1 = 0

(x – 1)(x – 1) = 0

x=1

Apabila x = 1, y = 13 – 3(1)2 + 6(1) + 2
=1–3+6+2
=6
º (1, 6)

7. (a) h(x) = kx3 – 4x2 – 5x

h(x) = 3kx2 – 8x – 5

(b) h(1) = 8

3k(1)2 – 8(1) – 5 = 8

3k – 13 = 8

3k = 21

k = 7
( ) 8. (a) y =
3 x – 1 4
4 6
[ ( ) ] ( )dy
3 x d x
dx 4 6 dx 6
= (4) – 1 4–1 –1

( ) ( )= 3x – 1 3 1
6 6

( )= 1 x –13
2 6

(b) y = 112(10x – 3)6
[ ]dy

dx
= 112(6)(10x – 3)6 – 1 d (10x – 3)
dx

= 1 (10x – 3)5(10)
2
= 5(10x – 3)5
(c) y = 8
2 – 5x

y = 8(2 – 5x)–1

dy = [8(–1)(2 – 5x)–1 – 1] d (2 – 5x)
dx dx
= –8(2 – 5x)–2(–5)

= (2 40
– 5x)2

( )(d) y =x– 1 3
x

= (x – x–1)3

dy = 3(x – x –1)3 – 1 d (x – x –1)
dx dx

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

(e) y = 1 9x
3! 3 –
= (3 – 9x)– 13

23

dy = – 31 (3 – 9x )– 31 –1 d (3 – 9x)
dx = – 31 (3 – 9x )– 43 dx

(–9)

= 3
4
(3 – 9x)3

= 3
3! (3 – 9x)4

(f) y = ! x2 + 6x + 6
1

= (x2 + 6x + 6)2

dy = 1 (x2 + 6x + 1 – 1 d (x2 + 6x + 6)
dx 2 dx
6)2

[ ] = 1 (x2 + 6x + 6)– 12 (2x + 6)
2
x+3
= ! x2 + 6x + 6

9. y = 24
(3x – 5)2

= 24(3x – 5)–2

dy = – 48(3x – 5)–2 – 1 d (3x – 5)
dx dx

= – 48(3x – 5)–3(3)

= – (3x14–45)3
dy
Apabila x = 2, dx = – (3(21)4–4 5)3

= –144
( ) 10.
d1 = d (3x – 2)–3
dx (3x – 2)3 dx

= –3(3x – 2)–4(3)
9
= – (3x – 2)4

Secara perbandingan, a = 9 dan b = 4

11. (a) d [4x(2x – 1)5] = 4x[5(2x – 1)4(2)] + (2x – 1)5(4)
dx = 40x(2x – 1)4 + 4(2x – 1)5

= 4(2x – 1)4[10x + (2x – 1)]

= 4(12x – 1)(2x – 1)4
d
(b) dx [x4(3x + 1)7] = x4[7(3x + 1)6(3)] + (3x + 1)7(4x3)
= 21x4(3x + 1)6 + 4x3(3x + 1)7

= x3(3x + 1)6[21x + 4(3x + 1)]

= x3(33x + 4)(3x + 1)6

3)– 21(1)
[ ]( )(c) dx! x + 3 =x 1 (x + + ! 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

24

(d) d [(x + 7)5(x – 5)3] = (x + 7)5[3(x – 5)2(1) + (x – 5)3[5(x + 7)4(1)]
dx = 3(x + 7)5(x – 5)2 + 5 (x + 7)4(x – 5)3

= (x + 7)4(x – 5)2[3(x + 7) + 5(x – 5)]

= (x + 7)4(x – 5)2(3x + 21 + 5x – 25)

= (x + 7)4(x – 5)2(8x – 4)

= 4(2x – 1) (x + 7)4(x – 5)2

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

– 1 2+! x! x – 1 – ! x
2! x
=
(1 + ! x )2
–1 – ! x – 1 + ! x
=
2! x (1 + ! x )2
2
= – 2! x ! x )2
(1 +
1
= – ! x (1 + ! x )2

1)– 12(4)
x [ ]( )! 4x + 1 (1) – x 1 (4x +
! 4x + 1 2
( )(f) d = (! 4x + 1)2
dx

! 4x + 1 – 2x 1
! 4x +
=
4x + 1

4x + 1 – 2x
! 4x + 1
= 4x + 1

= 2x + 1
(4x + 1)! 4x + 1

= 2x + 1
! (4x + 1)3

( )(g) d (x2 + 2x + 7)(0) – 1(2x + 2)
dx 1 = (x2 + 2x + 7)2
x2 + 2x + 7

= – (x22+(x2+x 1)
+ 7)2

1 – 2x3 (x – 1)(–6x2) – (1 – 2x3)(1)
x–1 (x – 1)2
( )(h) d =
dx

= –6x3 + 6x2 – 1 + 2x3
(x – 1)2

= 6x2 – 4x3 – 1
(x – 1)2
12. f (x) = x! x2 + 3
1
= x(x2 + 3)2
( )f (x) = (2x)(x2 + 3)– 21 1
x 1 + (1)(x2 +
2 3)2

= x2 + ! x2 + 3
! x2 + 3

= x2 + x2 + 3
! x2 + 3

= 2x2 + 3 (Tertunjuk)
! x2 + 3

25

13. y= 4x – 3
x2 + 1

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

= 4x2 + 4 – 8x2 + 6x
(x2 + 1)2

= 4 + 6x – 4x2
(x2 + 1)2

Untuk ddyx adalah positif, 42x+2 6x – 4x2 . 0
– 3x – 2 , 0

(2x + 1)(x – 2) , 0

– 12 , x , 2 …1

Untuk y adalah positif, 4x – 3 . 0

x . 3 …2
4
3
Gabungkan 1 dan 2, diperoleh 4 , x , 2.

14. y= x–2
x2 + 5

dy = x2 + 5 – 2x (x – 2)
dx (x2 + 5)2
x2 + 5 – 2x2 + 4x
= (x2 + 5)2

= 5 + 4x – x2
(x2 + 5)2

Untuk dy , 0,
dx

5 + 4x – x2 , 0

x2 – 4x – 5 . 0

(x + 1)(x – 5) . 0
x , –1 atau x . 5 …1
Untuk y , 0,
x – 2 , 0
x , 2 …2

Gabungkan 1 dan 2, diperoleh x , –1.

Kuiz Pantas (Halaman 50)

Diberi y = 5x – 3.

dy = 5
dx

( )(a) dy 2­= 52
dx
= 25
d 2y
(b) dx2 = 0

( ) Jadi, dy 2 ≠ d 2y .
dx dx2

Latihan Kendiri 2.6

1. (a) y = 3x4 – 5x2 + 2x – 1

dy = 12x3 – 10x + 2
dx

d 2y = 36x2 – 10
dx2

26

(b) y = 4x2 – 2
x

= 4x2 – 2x–1
dy
dx = 8x + 2x–2
= 8x + 2
x2

d 2y =8– 4x–3
dx2
4
= 8 – x3

(c) y = (3x + 2)8
dy
dx = 8(3x + 2)7(3)

= 24(3x + 2)7

d 2y = 24(7)(3x + 2)6(3)
dx2
= 504(3x + 2)6

2. (a) f (x) = ! x + 1
x2
1
= x2 + x –2

f (x) = 1 x– 21 – 2x–3
2
1 2
= 2! x – x3

f (x) = –  1 x– 32 + 6x–4
4
1 6
= –  3 + x4

4x2

(b) f (x) = x4 + 2
x2

= x2 + 2x–2

f (x) = 2x – 4x–3

= 2x – 4
x3

f (x) = 2 + 12x– 4

= 2 + 12
x4

(c) f(x) = 2x +5
x –1
(x – 1)(2) – (2x + 5)(1)
f (x) = (x – 1)2

= 2x – 2 – 2x – 5
(x – 1)2

= – (x 7 1)2
–

f (x) = (x – 1)2(0) – [(–7)(2)(x – 1)(1)]
[(x – 1)2]2

= 14(x – 1)
(x – 1)4

= 14
(x – 1)3

27