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1 (a) 4 (b) 8 (c) 16 (d) 2 (e) 32 (f) 8 (g) 2 2 (a) 1 536 (b) 7 (c) 448 (d) 3 584 (e) 4 3 (a) 4 (b) 150 (c) 20 (d) 10 (e) 625 (f) 75 (g) 30 (h) 200 (i) 175 (j) 3 (k) 20 4 (a) 5510 (b) 1210 (c) 2310 (d) 12310 (e) 4310 5 (a) 10000002

(b) 1010002 (c) 10100012 6

(a) (b)12510 17910 22010(c)as as

11111012 101100112 110111002

7 (a) 17010 (b) 5810

(c) 5010 (d) 31510

(e) 20710 8 (a) 308

(b) 738 (c) 2648 9

(a) (b)30010 51210 160910(c)as as

4548 10008 31118

10 (a) 7410

(b) 1810 (c) 5110 (d) 38810 (e) 19310 11 (a) 1215

BAB 1 Asas Nombor

1.1 Nombor dalam Asas Dua, Asas Lapan dan Asas Lima

BAB 2 Graf Fungsi II

2.1 Graf Fungsi

1 (a) x –4 –3 3

y –5 –3 9

y

x

y = 2x + 3

–5

1 2 3–1–2–3–4 0

–5

–10

–15

(b) x –3 –2 –1 0 1 2 3

y –37 –18 –5 2 3 –2 –13

(c) x –2 –1 0 1 2

y –25 –4 –1 2 23

y

x

y = 3x3 – 1

1 2–1–2 0

20

30

10

–10

–20

–30

y

x1 2 3–1–2–3 0

5

–5

–10

–15

–20

–25

–30

–35

–40

y = –3x2 + 4x + 2

JAWAPAN (b) 2005 (c) 10005 (d) 11025 (e) 12415 (f) 30245 12 (a) 158 (b) 258 (c) 678 (d) 1168 13 (a) 305 (b) 1105 (c) 2345 (d) 3235 14 (a) 11012 (b) 100112 (c) 1000002 (d) 1101112 15 (a) 235 (b) 1005 (c) 1035 (d) 1405 16 (a) 168 (b) 258 (c) 638 (d) 1368 17 (a) 101102 (b) 1110112

(c) 11101012 (d) 111001112 18 (a) 11012

(b) 100102 (c) 11102

(d) 1010012

(e) 101101219 (a) 1002

(b) 1102 (c) 10012

(d) 1112

(e) 10012

Latihan Bestari 1.1 1 50 2 448 3 37 4 1058 5 1005 6 1368 7 1000102

8 11102

SUDUT KBAT 1 (a) Q2 = 758 Q = 111101 (b) R10 = 758 R = 61 2 (a) 1m28 = 9010 m = 3 (b) 1m28 = 11010102 m = 5

PRAKTIS BAB 1Soalan Objektif 1 C 2 C 3 B 4 D 5 A 6 D 7 A 8 A 9 D 10 A11 D 12 C 13 B 14 A 15 C16 C

Anjakan Prima Math F5 Jaw 3rd.indd 1 9/10/2017 3:52:53 PM

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(d) x –3 –2 –1 0 1 2

y 59 21 7 5 3 –11

y

x1 2 3–1–2–3 0

50

60

30

20

10

–10

40

y = –2x3 + 5

(e) x –4 –3 –2 –1 1 2 3 4

y –7.5 –10 –15 30 30 15 10 7.5

y

x

y = 30

x

1 2 3–1–2–3 0 4–4

25

20

10

5

–5

–10

–15

–20

–25

–30

30

15

(f) x –4 –3 –2 –1 1 2 3 4

y 5.5 7.3 11 22 –22 –11 –7.3 –5.5

y

x

y = –22

x

1 2 3–1–2–3 0 4–4

25

20

10

5

–5

–10

–15

–20

–25

15

2 (a) (i) y = –14 (ii) x = –1.8, 2.3 (b) (i) y = 10 (ii) x = –1.5 (c) (i) y = 6 (ii) x = 1.8

3 (a) (c)

y

x10

45

y

x0

6

3

(b) (d)

y

x0

5

5

y

–3

10x

4 (a) y

x

5

2 5–1 0

x = 2 (b) y

x–1

–3

0

x =

14

14

32

(c)

y

x

5

–1 0

x =

34

34

52

5 (a) y

x0–3

–27

(b)

y

x0–2

24

6 (a) (b) y

x0

y

x0

(c)

y

x0

Latihan Bestari 2.1 1 (a) y = 2x2 + 1 Apabila x = –2, y = 2(–2)2 + 1 = 9 Apabila x = 2, y = 2(2)2 + 1 = 9

x –3 –2 –1 0 1 2 3

y 19 9 3 1 3 9 19

(b)

y

x

20

15

10

5

–5

0

6.5

y = 2x2 + 1

1 2 3–1–2–3–1.6 1.65 2.6

(c) (i) Apabila x = 2.6, y = 14. (ii) Apabila y = 6.5, x = –1.6 dan 1.65.

2 Untuk mencari pintasan-x, y = 0. x2 – 8x – 9 = 0 (x + 1)(x – 9) = 0 x = –1, x = 9 Untuk mencari pintasan-y, x = 0. [ y = 0 – 0 – 9 = –9 Untuk mencari paksi simetri,

x = – b2a

y

x–1 90–9

x = 4

4 = – –8

2(1) = 4

3 y + 1 = x3

y = x3 – 1 Untuk mencari pintasan-x, y = 0. [ x3 – 1 = 0 x = 3√1 = 1 Untuk mencari pintasan-y, x = 0. [ y = –1

y

x0

–11

4 2x + y = 5 Pintasan-x, y = 0. [ 2x + 0 = 5

x = 52

Untuk mencari pintasan-y, x = 0. [ y = 5

x

5

y

0 52

2.2 Penyelesaian Persamaan dengan Kaedah Graf 1 (a)

25

20

15

–5

1 2 3

y

x

y = x2 + 5x + 1

–4 0(0.8, 6)

y = –5x + 10

–1–2–3

10

5


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(b)

25

20

10

5

1 2 3–1–2–3

y

0

–5

–10

–15

–20

–25

x4–4

15

(1.1, –20)

y = –5x – 15

y = – 22x

(c)

20

30

10

1 2

y

y = 3x3 – 1

–40

–50

–60

–70

–80

(–2.8, –72)

y = 15x – 30

–10

–20

–30

–1–2–3 0x

–90

2 (a)

–5

1 2 3–1–2–3

y

x

y = 3x2 – 2x – 1

0

35

(1.75, 4.5)(–0.8, 2.2)

30

25

20

15

10

5y = x + 3

(b)

60

50

30

20

10

–1–2–3

y

y = –2x3 + 5

–10

–20

–30

–40

x

40

(–2.7, 47)

y = –10x + 20

0 1 2 3

↓

Latihan Bestari 2.2 1 (a) Graf fungsi: y = 2x2 + 3x – 1 ⇒ 2x2 = y – 3x + 1 Gantikan 2x2 = y – 3x + 1 ke dalam

2x2 + x – 6 = 0 y – 3x + 1 + x – 6 = 0 y = 2x + 5 (b) Graf fungsi: y = –3x2 – x + 7 ⇒ 3x2 = –y – x + 7

⇒ x2 = 13

(–y – x + 7)

Gantikan x2 = (–y – x + 7) ke dalam x2 + x + 5 = 0. x2 + x + 5 = 0 1

3(–y – x + 7) + x + 5 = 0

–y – x + 7 + 3x + 15 = 0 –y + 2x + 22 = 0 y = 2x + 22 (c) Graf fungsi: y = –2x3 + 5 ⇒ 2x3 = 5 – y Gantikan 2x3 = 5 – y ke dalam

2x3 + x + 7 = 0. 2x3 + x + 7 = 0 5 – y + x + 7 = 0 y = x + 12 (d) Graf fungsi: y = 3x3 + x – 7 3x3 = y – x + 7

x3 = 13

(y – x + 7)

Gantikan x3 = 13

(y – x + 7) ke dalam x3 – x + 1 = 0. x3 – x + 1 = 0

13

(y – x + 7) – x + 1 = 0

y – x + 7 – 3x + 3 = 0 y = 4x – 10 (e) 2x2 + 5x = 16

2x + 5 = 16x

2x + 5 = y

[ y = 2x + 5

2.3 Rantau yang Mewakili Ketaksamaan dalam Dua Pemboleh Ubah

1 (a) y

x0

y = –x + 2

(b) y

x0

y = 1

2x – 1

(c) (d)

y

x0

y =

2x +

4

y

x0

y =

5x –

6

2 (a) y x = 6

x + y = 6

0

y = x + 412

x

(b)

y

x0

y = x

+ 5

y = – 12 x

(c)

y

x0

y = 2x

– 6

y = –x + 2

y = 1

(d)

y

x0

y = 3x + 5

y = 5y = x

3 (a) x < 0 (b) x . –4 y > –x y >

12 x

y , 6 x + y < 0

(c) y < 6x x + y < 7 3y – x . 0 4 x < 0, y > –x –5 dan y < x + 4.

Latihan Bestari 2.3 1 y = 2x – 3 Apabila x = 0, y = 0 – 3 = –3. y

x

–3

10

–1

Apabila x = 1, y = 2 – 3 = –1. Pilih titik (0, 0). 0 2(0) – 3 0 > –3 Maka, titik (0, 0) memuaskan y > 2x – 3. 2 y = –x + 6 Apabila x = 0, y = 0 + 6 = 6. Apabila y = 0, x = 6. Pilih titik (0, 0). 0 0 + 6

y

x

6

0 6 0 , 6 Maka, titik (0, 0) memuaskan y , –x + 6. 3 Gantikan x = 2 dan y = 4 ke dalam ketaksamaan y . 2x – 9. y . 2x – 9 4 . 2(2) – 9 4 . 4 – 9 4 . –5 (Benar) Titik A(2, 4) berada dalam kawasan y . 2x – 9.


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4 Gantikan x = –4 dan y = 0 ke dalam ketaksamaan y . 4x + 1. y . 4x + 1 0 . 4(–4) + 1 0 . –16 + 1 0 . –15 (Benar) Titik B(–4, 0) berada dalam kawasan y . 4x + 1.

5 y > 0: Kawasan itu berada di atas garis = 0.

y < 12 x:

Kawasan itu berada di atas garis y = 12 x.

x < 6: Kawasan itu berada di atas garis x = 6.

x

y

y = 12

x

x = 6

0

SUDUT KBAT 1 (a) (9 – 2x)(13 – 2x) 117 – 18x – 26x + 4x2

4x2 – 44x + 117 (b) L = 4x2 – 44x + 117

x 0 1 2 3 4 5 6 7 8 9 10 y 117 77 45 21 5 –3 –3 5 21 45 77

y

1 2 30 4 5 6 7 8

90

100

110

80

60

50

40

20

10

–10

70

30

x

(c) (i) 2.2 m (ii) 1.5 m (d) 69 m2

PRAKTIS BAB 2Soalan Objektif 1 C 2 B 3 D 4 A 5 D 6 C 7 A 8 B 9 B 10 A11 D 12 C 13 A 14 B 15 B16 C 17 D 18 B 19 D 20 C

Soalan Subjektif

1 (a) Apabila x = –3, y = – 6(–3)

= 2

Apabila x = 0.5, y = – 60.5

= –12

(b)y

x

y = – 6x

1 2 3–1–2–3 0–4 4

12

10

6

4

2

–2

–4

–6

–8

–10

8

–12

y = –x – 2

(c) (i) y = –3.8 (ii) x = –0.9

(d) 6x = x + 2

⇒ –y = x + 2 y = –x – 2 Daripada graf, x = –3.6 dan 1.7. 2 (a) y = 2x2 – 3x – 2 k = 2(–3)2 – 3(–3) – 2 = 25 m = 2(2)2 – 3(2) – 2 = 0

(b) y

x1 2 3–1–2–3 0 4–4

45

40

30

25

20

10

5

35

–5

15

y = –2x + 28

y = 2x2 – 3x – 2

(c) (i) y = 18 (ii) x = –1.7, 3.2 (d) y = 2x2 – 3x – 2 ⇒ 2x2 = y + 3x + 2 Gantikan persamaan ke dalam 2x2 – x – 30 = 0. y + 3x + 2 – x – 30 = 0 y = –2x + 28 Daripada graf, x = –3.65. 3 (a) Apabila x = –2, y = (–2)2 – (–2) – 2 = 4 Apabila x = 0.5, y = (0.5)2 – (0.5) – 2 = –2.25

(b)y

x1 2 3–1–2–3 0 4

12

10

6

4

2

–2

–4

8

y = –2x + 8 y = x2 – x – 2

(c) (i) y = 3 (ii) x = –2.5, 3.45 (d) y = x2 – x – 2 ⇒ x2 = y + x + 2 Gantikan persamaan ke dalam x2 + x – 10 = 0. y + x + 2 + x – 10 = 0 y = –2x + 8 Daripada graf, x = 2.7.

BAB 3 Penjelmaan III

3.1 Gabungan Dua Penjelmaan

1 (a) 6

y

x

y = –1

AA�

C�

C��B��

A��

B�

B C

R

T4

2

–2

–4

–6

20–2–4–6 4 6

(b)

6

y

x

4

2

–2

–4

–6

20–2–4–6 4 6

B�

C�

y = –1

x = 1

A�

A�

B�

C�

BC

A

(c)

6

y

x

4

2

–2

–4

–6

20–2–4–6 4 6C��

B��

A��

D��

C�

D�

A�

B�

A

B

CD

y = –x


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(c) y

x

4

2

–2

–4

20–2–4–6–8 4 6C��

A��

B�

C�

A�

B

A

C

x = –1

(d) y

x

4

2

–2

–4

20–2 4 6 8

C��

B�� A��

C�

A�

B�AB C

(e) y

x

4

2

–2

–4

20–2–4–6–8 4 6A��

C��

B��

A� B�

C�

A B

C

(f)

6

y

x

4

2

–2

–4

420–2–4–6 6 8A��

B��

y =

x

C��C�B�

A�

BA

C

4 (a) (2, 4) (e) (3, 4) (b) (–6, –1) (f) (4, 2) (c) (6, –4) (g) (–2, 4) (d) (–4, –2) 5 (a) PQ tidak sama dengan QP (b) PQ tidak sama dengan QP

6 (a) (i) U ialah translasi 8–1 .

(ii) V ialah pantulan pada garis y = –1.

(b) (i) U ialah translasi –1 5 .

(ii) V ialah putaran 90° ikut arah lawan jam melalui (0, 0).

(c) (i) U ialah translasi 74 .

(ii) V ialah pembesaran dengan faktor skala 2 pada pusat (1, 1).

(d) (i) U ialah pantulan pada garis x = –1.

(ii) V ialah translasi –7–4 .

(e) (i) Q ialah pantulan pada garis x = –1. (ii) P ialah putaran 90° ikut arah jam

melalui (0, 0). 7 (a) Putaran 180° ikut arah jam melalui (0, 0). (b) Putaran 180° ikut arah jam melalui (0, 0). (c) Putaran 90° ikut arah lawan jam melalui (2, 4).

Latihan Bestari 3.1 1

6

y

x

A�A�� 4

2

–2–2 0–4 2 4

A

Koordinat imej = (–4, 4)

2

6

y

x

P��4

2

–2–2 0–4 2 4

P�P

Koordinat imej = (1, 5)

3

y

PQ

RC

B

A

x

8

6

4

2

2

(7, 9)

04 6 8 10 12

V ialah putaran 90° ikut arah jam melalui (7, 9).

SUDUT KBAT 1 (a) N adalah pembesaran pada pusat T

dengan faktor skala 12

. M adalah pantulan

pada garis US.

(b) QSU = 30 × 12 2

QSU = 7.5 cm2

Luas kawasan berlorek = 30 – 7.5 = 22.5 cm2

PRAKTIS BAB 3Soalan Objektif 1 C 2 B 3 C 4 D 5 B 6 C 7 A 8 C 9 B 10 C11 B 12 C 13 C 14 C 15 D16 D

Soalan Subjektif 1

P�

P��

P

10

y

x

y = 5

8

6

4

2

20

4 6 8 10 12

(d)

6

y

x

4

2

–2

–4

–6

20–2–4–6 4 6C��

B��

A��

C�

B�

A�

B

AC

(e)

6

y

x

4

2

–2

–4

–6

20–2–4–6 4 6

C�� A��

B��

A�

B�

C�

B

A

C

(f)

6

y

x

4

2

–2

–4

–6

20–2–4–6 4 6

C��

B��

A��

C�

B�

A�

B

C

A

2 (a) II (c) I (b) II (d) II 3 (a)

6

y

x

4

2

–2

–4

–6

20–2–4–6–8 4 6B

C

A

A��

C��

B��

C�

A� B�

(b)

6

y

x

4

2

–2

–4

–6

20–2–4–6–8 4 6BA

C

C�

A� B�

C��

A�� B��


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(a) A : Pantulan pada garis y = 5

(b) B : Translasi 54

2 y

x

x = 6

8

6

4

2

20

4 6 8 10 12 14

A�B�/B��

C�

A��

C��

A B

C

(a) Q : Pantulan pada garis x = 6 (b) P : Pembesaran dengan faktor skala 2 pada pusat (8, 3). 3 10

y

x

8

6

4

2

20

4 6 8

(5, 2)

90°

10 12 14

P��

P�P

(a) A : Putaran 90° ikut arah lawan jam (6, 5) (b) B : Pembesaran dengan faktor skala 3 pada pusat (5, 2). 4

y

x

8

6

4

2

2

(1, 1)0

4 6 8 10 12 14 16 18

P��

P�

P

(a) A : Pembesaran dengan faktor skala 2 pada pusat (1, 1).

(b) B : Translasi 7–2 .

5 Luas objek = 11 cm2

Faktor skala = 2 [ Luas imej = 22 3 11 cm2

= 4 3 11 cm2

= 44 cm2

6 Luas objek = 18 cm2

Faktor skala = 2 Luas imej = 22 3 18 cm2

= 72 cm2

[ Luas kawasan berlorek = 72 – 18 = 54 cm2

7 (a)

8

y

x

y = 5

6

4

2

0–2 42 8 10

(8, 7)

(6, 4)

(8, 5)

6 (i) (8, 7) (ii) (8, 5)

(b)

y

C BF G

E AD

N M

x

y = x + 1

K L

8

6

4

2

20

4 6 8 10 12 14 16

(i) V ialah pantulan pada garis y = x + 1. (ii) Putaran 90° ikut arah lawan jam melalui titik (6, 7). (c) (i)

10

y

x

N M

K Z

YX

L

8

6

4

2

20

4 6 8 10 12 14

(7, 1)

16

Pusat pembesaran = (7, 1) (ii) Luas imej = 530 cm2

Faktor skala = 2 Luas objek

= 530 cm2

(2)2 = 132.5 cm2

8 (a)

8

10

y

x

y = 3

6

4

2

–20–2 42 8 10

(8, 7)

(8, –1)

(5, 1)

6

(i) (8, –1) (ii) (8, 7)

(b) y

x

B C

A D

F

HE

G

8

6

4

2

20

4 6 8 10 12 14

U ialah putaran 90° ikut arah lawan jam melalui titik (6, 7). W ialah pembesaran dengan faktor

skala 2 pada pusat (7, 6). (c) Andaikan ABC = x cm2. Luas imej = 22 3 x 20 + x = 4x 3x = 20

x = 203

= 6 23

cm2

9 (a)

4

y

x

2

–4

–2

–2 0–4 2 4 6

M

M�

N��

N�

N2�

N2��

N

(i) Imej of M = (–2, 4)

(ii) (a) Imej of N = (–4, 2) (b) Imej of N = (1, 0) (b)

6

y

A

B

DC

FE

GH

M

L

J

K

x = –1

x

4

2

2–4 –2 0–6 2 4 6

(3, 4)

8

(i) (a) V ialah pantulan pada garis x = –1. (b) W ialah pembesaran dengan faktor skala 3 pada pusat (3, 4). (ii) Luas imej = 167.4 cm2

Faktor skala = 3 Luas objek = 167.4 cm2

32

= 18.6 cm2

[ Luas kawasan berlorek = 167.4 cm2 – 18.6 cm2

= 148.8 cm2

10 (a)

6

y

x

4

2

–2–2 0–4 2 4 6

(3, 0)

(0, 2)

(–2, –1)

(i) (–2, –1) (ii) (0, 2) (iii) (3, 0) (b)

y

G E D A B

C

O

F

x

x = 9

6

4

2

2 4 6 8 10 12 14 (i) U ialah pantulan pada garis x = 9. (ii) V ialah pembesaran dengan faktor skala 2 pada pusat (8, 6). (c) Luas objek = 16 cm2

Faktor skala = 2 Luas imej = 22 3 16 cm2

= 64 cm2

[ Luas kawasan berlorek = 64 cm2 – 16 cm2 = 48 cm2


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BAB 4 Matriks

4.1 Matriks 1 (a) 2; 3; 2 × 3 (d) 3; 2; 3 × 2 (b) 2; 2; 2 × 2 (e) 3; 3; 3 × 3 (c) 1; 3; 1 × 3 2 (a) 4 (d) 8 (b) 8 (e) –1 (c) 6 (f) 0

Latihan Bestari 4.1 1 2 × 1 2 1 × 1 3 7

4.2 Matriks Sama 1 (a) Sama (c) Tidak sama (b) Tidak sama (d) Sama 2 (a) p = 7, q = 2 (b) p = –7, q = –3

(c) p = 4, q = 32

(d) p = 1, q = 5

(e) p = 28, q = 7

Latihan Bestari 4.2 1 –5 3 p = 1; q = –3

2 p = 7; q = 6 4 p = –6; q = 12

4.3 Penambahan dan Penolakan Matriks

1 (a) 77 (e) 4

–5 (b) (5 12) (f) (1 –5 –11)

(c) 11 –2–4 4 (g) –3 1 –1

–1 –2 8

(d) 7 1 2 1212 –2 (h) 4 –7

10 –4

2 (a) –3 0 (c) 3 2

–10 7 (b) (–1 6) 3 (a) x = 8 (b) x = 5, y = 11 (c) x = –4, y = –10 (d) x = 2, y = 2 (e) x = 8, y = 3

Latihan Bestari 4.3 1 (2 6) + (–7 3) = (2 + (–7) 6 + 3) = (–5 9)

2 –3 6 1 7 – 5 2

–7 0 = –3 – 5 6 – 2 1 – (–7) 7 – 0

= –8 4 8 7

3 17 – 6

–3 + 51 = 1 – 6 + 5

7 – (–3) + 1 = 0

11 4 (–2 6) – (3 5) + (2 3) = (–2 – 3 + 2 6 – 5 + 3) = (–3 4)

5 43 – 2

x = 210

4 – 23 – x = 2

10 3 – x = 10 x = 3 – 10 = –7 6 x – 7 = 0 2 – 6 = 2y ⇒ x = 7 –4 = 2y y = –2

4.4 Pendaraban Matriks dengan Nombor

1 (a) 810 (d) 6

2 (b) –12

3 (e) –18 –1

(c) –217 (f) 2 8

–6 –12 2 (a) 1

3 (c) 2

23

(b) 114

Latihan Bestari 4.4

1 –2 6–4 4

2 –9 15 3 –6

3 –6 + 18 8 – 15 = 12

–7 4 1

2 (4 –8) – 3(1 –2) = (2 –4) – (3 –6) = (–1 2) 5 –8 + 3x = 6 ⇒ 3x = 14

x = 143

6 –12 y – 6 = 5 ⇒ –

12 y = 11

y = –22

4.5 Pendaraban Dua Matriks 1 (a) Ya, 1 × 1 (b) Ya, 2 × 2 (c) Ya, 2 × 1 (d) Tidak (e) Tidak (f) Ya, 3 × 3 2 (a) (10) (e) (0 2)

(b) 6 123 6 (f) 10 –7

10 –16

(c) –2 3 1–6 9 3 2 –3 –1 (g) (–3 9)

(d) –4–5

3 (a) 8 + 3x = 2 3x = –6 x = –2 (b) 3x = 6 x = 2 (c) 6 + 0 + x = 9 x = 3 (d) x = 5 (e) x = –2

(f) 6 – x = 2x 6 = 3x x = 2 (g) 6 – x = 3x 6 = 4x 3

2 = x

12 = y (h) –6 + x = –1 x = 5 3 + 3x = y 3 + 15 = y y = 18

Latihan Bestari 4.5 1 Tidak wujud.

2 –14 –10 21 15

3 –1 + 15 0 – 20 2 – 6 0 + 8 = 14 –20

–4 8 4 2x – 15 = 21 2x = 36 x = 18 5 5 – 6x = –7 –6x = –12 x = 2 6 –4 + 9w = 23 9w = 27 w = 3

4.6 Matriks Identiti 1 (a) B ialah matriks identiti kerana AB = A. (b) P ialah matriks identiti kerana PQ = Q. (c) x = 1

2 (a) 1 0 00 1 00 0 1

(b) x = 1 (c) p = 1, q = 0, r = 0, s = 1

3 (a) 2 7 53 –1 65 –3 1

(b) x = 7, y = 9 (c) x = 5

Latihan Bestari 4.6

1 2 1–5 4 1 0

0 1 = 2 1–5 4

1 00 1 ialah matriks identiti.

2 p 75 q = –2 7

5 9 [ p = –2, q = 9 ⇒ p + q = 7

3 x 00 x ialah matriks identiti, maka x = 1.

4 x 2y 7 = –1 2

–3 7 x = –1, y = –3 Maka, x + y = –1 – 3 = –4


J8

4.7 Matriks Songsang

1 (a) 1

(5 – 6)5 23 1 = –5 –2

–3 –1

(b) 1

(–8 – 6) 4 –2–3 –2 = – 1

14 –4 –2–3 –2

= – 27

17

3

14

17

(c)

1(–2 – 12) –2 –4

–3 1 = – 114 –2 –4

–3 1

= 17

27

3

14

– 114

(d)

1(–8 + 15) –4 5

–3 2 = 17 –4 5

–3 2

= – 47

57

37

27

–

(e) 1

(2 + 6) 2 2–3 1 = 1

8 2 2–3 1

= 14

14

38

18

–

2 (a) 1

(–8 + 9) 4 3–3 –2 = 1

1 4 3–3 –2

= 4 3–3 –2

q = –3 (b) m = 14 – 12 = 2 (c) 10 – 3p = 0 3p = 10 p = 10

3

(d) 12 + 4h = 0 4h = –12 h = –3

Latihan Bestari 4.7

1 120 – 18 5 2

9 4 = 52

1

92

2 2 A–1 = 1

–16 + 15 –4 3–5 4 = 4 –3

5 –4 3 A–1 = 1

21 – 20 –3 –4–5 –7 = –3 –4

–5 –7 [ p = –5

4 B = A–1 = 124 – 20 4 –5

–4 6

= 1 – 5

4

–1 3

2

5 Q = P–1 = 1–24 + 21 6 –3

7 –4

= –2 1

– 7

3 43

6 P = Q–1 = 1–28 + 30 –7 5

–6 4

= – 72

52

–3 2

4.8 Penyelesaian Persamaan Linear Serentak dengan Kaedah Matriks

1 (a) 2 –13 1 x

y = –1 6

(b) 3 –1–2 1 x

y = 5–3

(c) 2 00 3 x

y = 74

2 (a) 2 –1–1 3 x

y = 7–11

xy =

1(6 – 1)

3 11 2 7

–11

= 15 21 – 11

7 – 22

= 2–3

\ x = 2, y = –3

(b) 2 11 –3 x

y = 11–5

xy =

1(–6 – 1)

–3 –1–1 2 11

–5

= – 17 –33 + 5

–11 – 10

= 43

\ x = 4, y = 3

(c) 5 22 –1 x

y = –4–7

xy =

1(–5 – 4) –1 –2

–2 5 –4–7

= – 19 4 + 14

8 – 35

= –23

\ x = –2, y = 3

(d) 5 23 –1 x

y = –7–13

xy =

1(–5 – 6)

–1 –2–3 5 –7

–13

= – 111

7 + 2621 – 65

= –34

\ x = –3, y = 4

(e) 2 33 –1 x

y = –11–11

xy =

1(–2 – 9)

–1 –3–3 2 –11

–11

= – 111

11 + 3333 – 22 = –4

–1 \ x = –4, y = –1

Latihan Bestari 4.8

1 2 –13 1 x

y = 61

2 –1 5 6 –3 x

y = 2–5

3 2 1–3 2 x

y = –511

xy = 1

4 + 3 2 –13 2 –5

11 = 1

7 –10 – 11–15 + 22

= –3 1

[ x = –3, y = 1

4 4 –3–3 1 x

y = 5–5

xy = 1

4 – 9 1 33 4 5

–5 = 1

–5 5 – 1515 – 20

= 21

[ x = 2, y = 1

SUDUT KBAT 1 (a) P + Q = 230 8P + 3Q = 1 290

1 18 3 P

Q = 2301 290

(b) 1 18 3 P

Q = 2301 290

PQ = 1 1

8 3 –1

2301 290

= 1(1)(3) – (1)(8)

3 –1–8 1 230

1 290 = – 1

5 (3)(230) + (–1)(1 290)(–8)(230) + (1)(1 290)

= – 15 –600

–550 P

Q = 120110

P = 120 Q = 110 P – Q = 120 – 110 = 10


J9

PRAKTIS BAB 4Soalan Objektif 1 A 2 C 3 B 4 B 5 D 6 B 7 A 8 C 9 D 10 C11 D 12 B 13 D 14 A 15 A16 A

Soalan Subjektif

1 (a) M = 12 + 15 1 3

–5 2 =

117

317

– 5

17

217

(b) 2 –3

5 1 xy = –7

8 x

y = 1

17

317

– 5

17

217

–7 8

= 13

[ x = 1, y = 3

2 (a) 2 31 –2

–1 = 1

–4 – 3 –2 –3–1 2

= – 17 –2 –3

–1 2 = m –2 p

–1 2

[ m = –17 , p = –3

(b) 2 31 –2 x

y = –1–4

xy = – 1

7 –2 –3–1 2 –1

–4 = – 1

7 2 + 121 – 8

= –2 1

[ x = –2, y = 1

3 (a) Q = P–1

= 1

4 + 3 2 3–1 2

= 17 2 3

–1 2 = k 2 h

–1 2 [ k =

17 , h = 3

(b) 2 –31 2 x

y = –110

xy = 1

7 2 3–1 2 –1

10 = 1

7 –2 + 301 + 20

= 43

[ x = 4, y = 3

4 (a) 3 1–1 2

–1 = 1

6 + 1 2 –11 3

= 27

– 17

17

37

= 27

– 17

17

h [ h = 3

7

(b) 3 1–1 2 x

y = 5–4

xy = 1

7 2 –11 3 5

–4 = 1

7 10 + 45 – 12= 2

–1 [ x = 2, y = –1

5 (a) Q = 1–3 – 8 –3 –4

–2 1 =

311

411

2

11

– 111

(b) 1 4

2 –3 xy = 1

13

xy = – 1

11 –3 –4–2 1 1

13 = – 1

11 –3 – 52–2 + 13

= 5–1

[ x = 5, y = –1

6 (a) 1 24 5

–1 = 1

5 – 8 5 –2–4 1

= – 13 5 –2

–4 1 = n 5 q

–4 1

[ n = –13 , q = –2

(b) 1 24 5 u

v = –1–7

uv = – 1

3 5 –2–4 1 –1

–7 = – 1

3 –5 + 144 – 7

= – 13 9

–3 = –3 1

[ u = –3, v = 1

7 (a) R = P –1

= 1–2 – 3 –1 –3

–1 2

= – 15 –1 –3

–1 2 = k –1 h–1 2

[ k = –15 , h = –3

(b) 2 31 –1 u

v = –6 7

uv = – 1

5 –1 –3–1 2 –6

7 = – 1

5 6 – 216 + 14

= – 15 –15

20 = 3–4

[ u = 3, v = –4

8 (a) 4 –32 1

–1 = 1

4 + 6 1 3–2 4

= 110 1 3

–2 4

= 1

10

310

– 1

5

25

[ p =

25

(b) 4 –32 1 u

v = 1–7

uv = 1

10 1 3–2 4 1

–7 = 1

10 1 – 21–2 – 28

= –2–3

[ u = –2, v = –3

9 (a) R = 1–2 – 3 –2 –1

–3 1

= – 15 –2 –1

–3 1 =

25

15

35

– 15

(b) 1 13 –2 u

v = –4–7

uv = – 1

5 –2 –1–3 1 –4

–7 = – 1

5 8 + 712 – 7

= –3–1

[ u = –3, v = –1

10 (a) 1 15 –2

–1 = 1

–2 – 5 –2 –1–5 1

= – 17 –2 –1

–5 1 = p –2 –1

–5 q [ p = –

17 , q = 1

(b) 1 15 –2 x

y = 1–16

xy = – 1

7 –2 –1–5 1 1

–16


J10

= – 17 –2 + 16

–5 – 16 = – 1

7 14–21 = –2

3

[ x = –2, y = 3

BAB 5 Ubahan

5.1 Ubahan Langsung

1 (a) Ya (b) Tidak (c) Ya (d) Tidak (e) Tidak 2 (a) y = kx 2 = k(8) k = 1

4

y = x4

(b) k = 51

y = 5√x

(c) k = 510

= 12

y = √x2

(d) k = 5025

= 2

y = 2x2

(e) k = 168

= 2

y = 2x

3 (a) k = 305

= 6

y = 6x

y = 6(7) = 42

(b) k = 510

= 1

2

n = 12

(7)

= 3 . 5

(c) y = kx2

43

= k(4)

k = 1

3

y = 13

(25)

= 25

3

(d) y = kx2

92

= k(9)

k = 1

2

w = 12

(16)

= 8

(e) y = kx3

16 = k(8) k = 2 y = 2(125) = 250

(f) p = k√ t 5 = k√100

k = 1

2

3 = 12

√n

n = 62

= 36

(g) p = kq2

43

= k(4)

k = 1

3

3 = 13

q2

q2 = 9

q = 3

(h) u = k√ v 2 = k√100

k = 1

5

5 = 15

√ n

√ n = 25 n = 625

Latihan Bestari 5.1 1 p x2

⇒ p = kx2

p1

x12 =

p2

x22

2422

= p42

244

= p

16

p = 96

2 y x3

⇒ y = kx3

y1

x13 =

y2

x23

4523

=

100x3

x3 = 1 000 x = 10

3 p √q ⇒ p = k√q p1

√q1

= p2

√q2

1√49

= w√81

w = 97

5.2 Ubahan Songsang 1 (a) Ya (b) Tidak (c) Ya

(d) Tidak (e) Tidak 2 (a) k = 128 p = 128

q2

(b) k = 30 × 2 = 60 y = 60

√ x (c) k = 21 × 4 = 84 y = 84

√ x (d) k = –36 × 2 = –72 p = – 72

q

(e) k = –648 P = – 648

q3 3 (a) k = 9 × 4 = 36 y = 36

72

y = 1

2

(b) k = 72 × 4 = 288 y = 288

9 = 32

(c) k = 18 × 16 = 288 p = 288

36

= 8 (d) k = 80 × 4 = 320 y = 320

10

= 32 (e) k = 80 × 1

8

= 10 640 = 10

q3

q3 = 1

64

q = 1

4

Latihan Bestari 5.2

1 y = 1x2

⇒ y = kx2

x12 y1 = x2

2 y2

4218 = 82y

y = 2

64

= 1

32


J11

2 y 1x3

⇒ y = kx3

x13 y1 = x2

3 y2 23(54) = x3(16)

x3 = 43216

= 27

x = 3

3 p 1√q

⇒ p = k√q

p1√q1 = p2√q2

n√64 = 16

(√144)

8n = 2

n = 14

5.3 Ubahan Tercantum

1 (a) y rt

(b) p 1qt

(c) p q2

t 2 (a) y = k

r2t

– 1

6 = k4(3)

6k = –12 k = –2 y = –2

r2t

(b) y = kq2r

k = –279

= –3 y = –3q2r

(c) y = k√ rt

1 =

k√ 48

2k = 8 k = 4 y = 4√ r

t

(d) p = kt√ r

– 12 = k

–4 × 4

2k = 16 k = 8 p = 8

t√ r

(e) p = kqx3

– 5 = k

1 × 8

4 4k = –40 k = –10 p = –10

qx3

(f) y = kr3

p –2 = k(–8)

2 –4 = –8k k =

12

y = r3

2p 3 (a) –1 = k(–2)

5

2k = 5 k = 5

2

n =

52 (4)

5 = 2 (b) k = –36

6 = –6 y = –6pq 30 = –6(–4)(w) 24w = 30 w = 5

4

(c) 1 = k(4)8

4k = 8 k = 2 6 = 2(9)

w 6w = 18 w = 3 (d) 1

2 =

k(3)18

6k = 18 k = 3 3 =

3(10)n

3n = 30 n = 10 (e) –8 = k12 (8) 4k = –8 k = –2 2 = –2(w)(4) –8w = 2 w = – 1

4

(f) –

=

1

2 k

112

(–8)

k = 13

w = 1

319

(27)

= 1

9

Latihan Bestari 5.3

1 y √rq

y = k√rq

2 = k√166

k = 3

[ y = 3√r

q

2 y p2

q y = kp2

q 1

3 = k 3 22

36 k = 3

[ y = 3p2

q

3 p

1qx3

p = kqx3

23

= k1

24 3 23

k = 29

[ p = 2

9qx3

4 p 1rq

p = krq

p1 q1 r1 = p2 q2 r2

–2 23

3 3 3 14

= 112

3 6 3 q

q = –4

5 y √rp

y1p1

√r1

= y2p2

√r2

(–2) 3 10√100

= 3 3 4√r1

√r = –6 r = 36

SUDUT KBAT

1 Masa a KeluasanPekerja

2 = k(200)8

k = 0.08

(a) t = 0.08(400)8

t = 4 hari (b) 4 = 0.08(L)

10

L = 500 Luas = 500 m2

PRAKTIS BAB 5Soalan Objektif 1 B 2 A 3 D 4 B 5 A 6 B 7 C 8 D 9 C 10 C11 D 12 B 13 C 14 C 15 C16 A 17 B 18 B 19 A 20 A


J12

BAB 6 Kecerunan dan Luas di bawah Graf

6.1 Kuantiti yang diwakili oleh Kecerunan Graf 1 (a) Kadar perubahan jarak melawan masa

atau laju dalam m s–1. (b) Kadar perubahan laju melawan masa atau

pecutan dalam km j–2.

2 (a) Jarak (km)

Masa(jam)

24

1620

84

12

10

2 3

(b) Jarak (m)

Masa (s)

6

45

21

3

10

2 3 3 (a) Kecerunan OP = 3 m s–1

Kecerunan PQ = 5 m s–1

Objek bergerak dengan laju seragam 3 m s–1 pada lapan saat pertama. Kemudian, ia menambah kelajuan kepada 5 m s–1 dan 7 saat lagi diperlukan untuk objek tiba di titik B. (b) Kecerunan OP = 80 km j–1

Kecerunan PQ = 0 Kecerunan QR = –64 km j–1

Bas itu bergerak dari Bandar A ke Bandar B dengan laju seragam 80 km j–1.

Kemudian, bas itu berhenti selama setengah jam. Selepas itu, bas itu bergerak

semula dari B ke A dengan laju seragam 64 km j–1.

4 (a) Laju = 124

= 3 m s–1

(b) Laju = 20 – 86

= 2 m s–1

(c) 8 – 2 = 6 saat (d) 0 m s–1 (kerana zarah ketika itu adalah

pegun) (e) Laju = 18

6

= 3 m s–1

(f) Laju = 248

= 3 m s–1

5 (a)

0g

2

N(RM)

2

12

Kadar satu pusingan ialah RM5. Caj permulaan ialah RM2.

(b) Isi padu (cm3)

0 10

80

Masa (s)

Isipadu (cm3)

Air mengalir pada kadar 8 cm3 sesaat.

Latihan Bestari 6.1 1 Kuantiti yang diwakili oleh kecerunan ialah

faedah, dalam RM, yang dikumpulkan daripada akaun deposit tetap yang selari dengan masa dalam tahun.

2 Kecerunan OP = 52 – 01 – 0

= 52 km j–1

Kecerunan PQ = 52 – 521.5 – 1

= 0 km j–1

Kecerunan QR = 0 – 522 – 1.5

= –104 km j–1

Bas itu bergerak dari bandar X ke bandar Y dengan laju purata 52 km j–1. Kemudian bas itu berhenti selama setengah jam. Kemudian, bas itu bergerak dari bandar Y ke bandar X dengan laju seragam 104 km j–1.

3 (a) Laju = Kecerunan

= 32 – 205 – 0

= 2.4 m s–1

(b) Laju = Kecerunan

= 0 – 3217 – 11

= –5 13

m s–1

4 (a) Tempoh = 12 s – 5 s = 7 saat (b) Laju = Kecerunan

= 0 – 2017 – 12

= –4 m s–1

(Laju negatif menandakan bahawa objek itu bergerak ke titik rujukan.)

6.2 Kuantiti yang diwakili oleh Luas di bawah Graf 1 (a) Harga yang perlu dibayar oleh pelanggan bagi mangga yang dibeli. (b) Jarak yang dilalui oleh zarah yang bergerak. 2 (a) Luas A + Luas B = 1

2 × (15 + 8) × 6 + 1

2 × 15 × 4

= 23 × 3 + 15 × 2 = (69 + 30) unit2

= 99 unit2

(b) Luas A + Luas B + Luas C = 1

2 × 8 × 10 + 10 × 7 + 1

2 × (20 + 10) × 6

= 40 + 70 + 90 = 200 unit2

3 (a) Jarak = 12

× 18 × 34 = 306 m

(b) Jarak = 12

× 25 × 8 = 100 m (c) Jarak = 30 × 10 = 300 m

(d) Jumlah jarak = 1

2 × (30 + 18) × 10 + (14 × 18)

= 240 + 252 = 492 m

(e) Jarak = (15 × 10) + 12

× (35 + 15) × 5 = 150 + 25 × 5 = 275 m (f) Jumlah jarak = 36 × 20 + 1

2 × 36 × 5

= 720 + 90 = 810 m

Purata laju = 81025

= 32.4 m s–1

(g) Jumlah jarak = 1

2 × (28 + 12) × 10 + 10 × 12 + 1

2

× (20 + 12) × 6 = 200 + 120 + 96 = 416 Purata laju = 416

26 = 16 m s–1

(h) Jumlah jarak = 1

2 × 10 × 50 + 1

2 × (50 + 28) × 4 + 28 × 6

= 250 + 156 + 168 = 574 Purata laju = 574

20 = 28.7 m s–1

(i) Jumlah jarak = 40 × 8 + 1

2 × (40 + 15) × 6 + 1

2

× (30 + 15) × 6 = 320 + 165 + 135 = 620 Purata laju = 620

20 = 31 m s–1

4 (a) 14 – x4

= 2 14 – x = 8 x = 6 (b) 0 – x

3 = –21

–x = –63 x = 63 (c) 20 – 8

t = 4

12 = 4t t = 3 (d) 0 – 24

t – 10 = –3

–24 = –3t + 30 3t = 54 t = 18 (e) v – 30

4 = –6

v – 30 = –24 v = 6


J13

(f) 0 – 3220 – t

= –4 –32 = –80 + 4t 4t = –32 + 80 4t = 48 t = 12 (g) 0 – 20

34 – t = –4

–20 = –136 + 4t 4t = 116 t = 29

Latihan Bestari 6.2 1 Isi padu petrol dalam liter 2

0

8

20

6 15 18x

A B

y

Luas di bawah graf = Luas A + Luas B = 1

2 3 (8 + 20) 3 6 + 1

2 3 (9 + 12) 3 20

= 84 + 210 = 294 unit2

3 Jumlah jarak = Luas di bawah graf = 1

2 3 (50 + 20) 3 12 +

12

3 (25 – 12) 3 20 = 420 + 130 = 550 m 4 Laju = Kecerunan

3 = 42 – x12 – 0

36 = 42 – x x = 6 m 5 Kadar perubahan laju = Kecerunan

4 = 40 – 12t – 0

4t = 28 t = 7 s 6 Jumlah jarak = Luas di bawah graf 426 = 1

2 3 (12 + 36) 3 t

+ 12

3 (20 – t) 3 36 426 = 24t + (360 – 18t) 426 = 6t + 360 t = 11 s

SUDUT KBAT 1 (a) 1210 – 1145 = 25 minit

(b) Laju = 60 km34

j

= 80 km j–1

Laju = 90 km

1.5 j = 60 km j–1

(c)

Masa

11000

20

40

60

80

100

1145 1210 1340

Laju (km j–1)

PRAKTIS BAB 6Soalan Subjektif 1 (a) Tempoh masa = 15 s – 9 s = 6 saat (b) Kadar perubahan laju = Kecerunan

= 0 – 3020 – 15

= –6 m s–2

(Tanda negatif menunjukkan bahawa ia adalah nyahpecutan) (c) Jumlah jarak yang dilalui = 471 m

12

3 (v + 30) 3 9 + = 471

12

3 (6 + 11) 3 30 9v + 270 + 510 = 942 9v = 162 v = 18 m s–1

2 (a) Tempoh masa = 1.0 jam – 0.6 jam = 0.4 jam (b) Purata laju

= Jumlah jarak yang dilaluiJumlah masa yang diambil

= 102 km3 jam

= 34 km j–1

(c) Graf ABCD menyilang graf AE pada ketika 0.9 jam dan jaraknya ialah 60 km.

(i) Maka, kereta dan bas berjumpa pada jarak 60 km. (ii) Bas itu memerlukan 0.9 jam untuk tiba di lokasi. 3 (a) Tempoh masa = 15 s – 6 s = 9 saat (b) Kadar perubahan laju = Kecerunan

= 12 – 86 – 0

= 46

= 23

m s–2

(c) Jumlah jarak yang dilalui = 222 1

2 3 (8 + 12) 3 6 +

[(15 – 6) 3 12] + = 222 1

2 3 (12 + 24) 3 (t – 15)

60 + 108 + (18t – 270) = 222 18t = 324 t = 18 s

4 (a) Laju seragam = 18 m s–1

(b) (i) Jarak yang dilalui = 144 (t – 16) 3 18 = 144 t – 16 = 8 t = 24 s (ii) Purata laju


12

3 (24 + 18) 3 16 + [144] +

=

12

3 (30 – 24) 3 18 30 = 336 + 144 + 54

30 = 17.8 m s–1

5 (a) Tempoh masa = 20 s – 12 s = 8 saat (b) Kadar perubahan laju = Kecerunan

= 46 – 1012 – 0

= 3 m s–2

(c) Jumlah jarak yang dilalui = 796 m 1

2 3 (10 + 46) 3 12 +

[8 3 46] + = 796

12

3 (t – 20) 3 46 336 + 368 + 23t – 460 = 796 23t = 552 t = 24 s 6 (a) Laju seragam = 20 m s–1

(b) (i) Jumlah jarak = 160 (t – 11) 3 20 = 160 t – 11 = 8 t = 19 s (ii) Purata laju


12

3 (8 + 19) 3 20 +

=

12

3 (20 + 35) 3 6 25 = 270 + 165

25

= 17.4 m s–1


= 0 – 2825 – 18

= –4 m s–2

(c) Jumlah jarak yang dilalui = 400 m

12

3 (v + 28) 3 13 + [(18 – 13) 3 28]

= 400

132

v + 182 + 140 = 400

v = 78 3 213

= 12 m s–1


J14

BAB 7 Kebarangkalian

7.1 Kebarangkalian Suatu Peristiwa 1 (a) S = {kepala, ekor} (b) S = {7, 8, 9, 10} (c) S = {k, k, b, b, b, b} (d) S = {P, Q, R, S, T} (e) S = {m, b, b, u, u, u} (f) S = {1, 1, 5, 5, 5, 5, 5, 10}

2 (a) 12

(d) 15

(b) 16

(e) 15

(c) 25

(f) 25

3 (a) 13

= 21

21 + x

21 + x = 63 x = 63 – 21 = 42 biji

(b) 12

= 1016 + x

16 + x = 20 x = 4

(c) 310

= 1218 + x

54 + 3x = 120 3x = 66 x = 22

(d) 25

× 20 = 8

(e) 12

= 1527 + x

27 + x = 30 x = 3

(f) 37

= 2130 + x

90 + 3x = 147 3x = 57 x = 19

(g) 13

= 1319 + x

19 + x = 39 x = 20

(h) 516

= 4065 + x

325 + 5x = 640 5x = 315 x = 63

Latihan Bestari 7.1 1 (a) {10, 11, 12, 13, 14, 15, 16, 17, 18, 19} (b) Nombor perdana = 11, 13, 17, 19

P(Nombor perdana) = 410

= 25

2 (a) {C, H, O, C, O, L, A, T, E} (b) Vowels = O, O, A, E

P(Vowels) = 49

3 P(durian) = 1812 + 18 + x

13

= 1812 + 18 + x

30 + x = 54

x = 24

7.2 Kebarangkalian Pelengkap Suatu Peristiwa 1 (a) (i) T9 ialah peristiwa mendapat nombor yang lebih besar daripada 2. (ii) T9 = {3, 4, 5, 6} (b) (i) P9 ialah peristiwa mendapat nombor yang bukan nombor perdana. (ii) P9 = {1, 4, 6} (c) (i) G9 ialah peristiwa memilih pelajar

lelaki. (ii) G9 = {lelaki} (d) (i) V9 ialah peristiwa memilih kad huruf konsonan. (ii) V9 = {B, C, D, F, G, H, J} (e) (i) A9 ialah peristiwa memilih oren. (ii) A9 = {oren}

2 (a) 1 – 47

= 37

(b) 1 – 38

= 58

(c) 1 – 710

= 310

(d) 1 – 45 = 1

5

(e) 1 – 5

9 = 49

Latihan Bestari 7.2 1 (a) Andaikan B ialah peristiwa memilih sebiji guli biru. [ B9 ialah peristiwa memilih guli hijau. (b) P(B9) = 1 – P(B)

= 1 – 411

= 7

11 2 (a) Andaikan A ialah peristiwa mengambil nombor perdana. [ A9 ialah peristiwa mengambil nombor bukan nombor perdana. (b) A9 = 1 – P(A)

= 1 – 29

= 7

9

7.3 Kebarangkalian Peristiwa Bergabung

1 (a) (i) A B = {2, 3, 4, 6} (ii) A B = {6} (b) (i) A B = {1, 2, 3, 5} (ii) A B = {3, 5}

2 (a) 59

(c) 136

(b) 14

8 (a) Tempoh masa = 1.8 jam – 0.9 jam = 0.9 jam (b) Purata laju = 140 km

2.5 jam = 56 km j–1

(c) Graf ABCD dan AEF bersilang pada jarak 110 km. Maka, teksi dan bas berjumpa pada jarak 110 km dari bandar R. 9 (a) Laju seragam = 12 m s–1

(b) (i) Jarak yang dilalui = 108 m (t – 8) 3 12 = 108 t – 8 = 9 t = 17 s (ii) Purata laju


12

3 (20 + 12) 3 8 + 108 +

=

12

3 (12 + 18) 3 (20 – 17) 20

= 128 + 108 + 4520

= 14.05 m s–1


= 15 – 06 – 0

= 2.5 m s–2

(c) Jumlah jarak yang dilalui = 295 m 1

2 3 (12 + 18) 3 15 +

= 295 1

2 3 (15 + v) 3 (22 – 18)

225 + 30 + 2v = 295 2v = 40 v = 20 m s–1

11 (a) Laju seragam = 16 m s–1

(b) (i) Jumlah jarak dilalui = 208 (28 – t) 3 16 = 208 28 – t = 13 t = 15 s (ii) Purata laju

12

3 (10 + 20) 3 9 +

= 1

2 3 (20 + 16) 3 (15 – 9)

+ 208 28

= 135 + 108 + 20828

= 16.1071

= 16.11 m s–1


J15

3 (a) 12

(e) 49

(b) 1

3 (f) 245

354

(c) 4

55 (g) 3

5

(d) 1

11

Latihan Bestari 7.3 1 S = {21, 22, 23, 24, 25, 26, 27, 28, 29, 30} P = {23, 29} Q = {22, 24, 26, 28, 30} P Q = {22, 23, 24, 26, 28, 29, 30} 2 S = {21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32,

33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50}

n(S) = 30 A = {25, 30, 35, 40, 45, 50} B = {21, 24, 27, 30, 33, 36, 39, 42, 45, 48} (a) A B = {21, 24, 25, 27, 30, 33, 35, 36, 39, 40, 42, 45, 48, 50} n(A B) = 14

[ P(A B) = 1430

= 715

(b) A B = {30, 45} n(A B) = 2

[ P(A B) = 230

= 115

3 = P[(B1 B2) (M1 M2)] = P(B1 B2) + P(M1 M2)

= 28

3 17 + 6

8 3 5

7 = 1

28 + 15

28

= 4

7 4

Kotak Bolamerah

Bolabiru Jumlah

A 2 4 6

B 3 7 10

= P[(RA LB) (LA RB)] = P(RA LB) + P(LA RB)

= 26

3 710 + 4

6 3 3

10 = 7

30 + 1

5

= 13

30 SUDUT KBAT

1 (i) 1260

= 15

(ii) 3260

3 3159

= 248885

(iii) 1260

3 1159 + 10

60 3 9

59 + 660

3 559

= 11295

+ 3118

+ 1118

= 21295

PRAKTIS BAB 7Soalan Objektif 1 A 2 B 3 B 4 D 5 C 6 C 7 D 8 A 9 A 10 B11 C 12 B 13 D 14 A 15 A16 C 17 A 18 A 19 D 20 C

Soalan Subjektif 1 (a) Andaikan F ialah peristiwa penumpang perempuan turun.

P(F1 F2) = 1220

3 1119

= 33

95 (b) P(sekurang-kurangnya seorang lelaki) = 1 – P(semua perempuan)

= 1 – 1118

3 1017

= 98153

2 (a) P(RM5 RM5) = 612

3 812

= 13

(b) P(RM1 RM1) + P(RM1 RM5) + P(RM5 RM1)

= 212

3 112 + 2

12 3 8

12 + 612

3 112

= 16

3 (a) P(M1 M2) = 68

3 57

= 1528

(b) P(M1 M2) + P(F1 F2)

= 46

3 35 + 2

6 3 1

5 = 7

15 4 (a) P(M G) = 6

18 3 4

17 = 4

51 (b) P(M1 M2) + P(G1 G2) + P(S1 S2)

= 618

3 517 + 4

18 3 3

17 + 818

3 717

= 49153

5 (a) P(T A) = 211

3 210

= 255

(b) P(M1 M2) + P(A1 A2) + P(T1 T2)

= 211

3 110 + 2

11 3 1

10 + 211

3 110

= 355

6 (a) P(A1 A2) = 420

3 319

= 3

95 (b) P(C A) = 2

20 3 4

19

= 2

95 7 (a) P(R1 B2) = 8

20 3 10

19

= 4

19 (b) P(R1 R2) + P(B1 B2) + P(G1 G2)

= 820

3 719 + 10

20 3 9

19 + 220

3 119

= 3795

8 (a) P(RA BB)

= 812

3 1012

= 59

(b) P(RA BB) + P(BA RB)

= 59

+ 4

12 3 2

12

= 1118

9 (a) P(B R) = 616

3 615

= 3

20 (b) P(B1 B2) + P(R1 R2)

= 416

3 315 + 2

16 3 1

15 = 7

120

10 (a) P(Bb Gc) = 410

3 810

= 825

(b) P(Bb Bc) + P(Gb Gc)

= 410

3 210 + 6

10 3 8

10 = 14

25

11 (a) P(VA CB) = 37

3 24

= 314

(b) P(VA VB) = 37

3 24

= 3

14 12 (a) P(BQ WP) = 3

18 + 2

18

= 5

18 (b) P(W W) + P(B B)

= 618

3 517 + 12

18 3 11

17 = 9

17


J16

13 (a) Andaikan R ialah peristiwa di mana pelanggan membayar dengan kad kredit dan S peristiwa di mana pelanggan

membayar secara tunai.

P(AR BS) = 12

3 25

= 1

5 (b) P(AS BS) + P(AR BR)

= 13

3 25 + 1

2 3 1

6 = 13

60 14 (a) P(Y Y) + P(R R)

= 412

3 311 + 8

12 3 7

11 = 17

33 (b) 1 – P(R R)

= 1 – 812

3 711

= 19

33 15 (a) P(E E)

= 49

3 38

= 16

(b) 1 – P(E E)

= 1 – 16

= 5

6

SUDUT KBAT 1 Bearing N dari M = 360° – 90° – tan–1 60

40 = 269.02°

U

N

R(20 × 2)

S(30 × 2)

M

PRAKTIS BAB 8Soalan Objektif 1 A 2 A 3 C 4 D 5 D 6 D 7 C 8 A 9 C 10 D11 A 12 A 13 C 14 C 15 C16 D 17 C 18 B 19 B 20 C

BAB 9 Bumi sebagai Sfera

9.1 Longitud 1 (a) U

S

P

(b) U

S

Q

2 (a) 110°T (b) 80°B (c) 50°T (d) 140°B (e) 80°T (f) 170°B (g) 130°T (h) 94°B 3 (a)

70°

0°

U

S

O

70°TMeridianGreenwich

(b)

0°110°

U

S

O

MeridianGreenwich

110°B 4 (a) 50° + 30° = 80° (b) 70° – 20° = 50° (c) 17° + 76° = 93° (d) 71° 179 – 39° 269 = 31° 519

(e) 26° 189 + 7° 419 = 33° 599

(c) U

C

B

200°

(d) N

B

C

300°

(e) U

Y

X

255.3°

(f) U

283.9°X

Y

4 (a) Bearing B dari A ialah 040°. (b) Bearing A dari B ialah 105°. (c) Bearing P dari Q ialah 210°. (d) Bearing Q dari P ialah 320°. (e) Bearing B dari A ialah 065°. (f) Bearing D dari C ialah 108°. (g) Bearing E dari F ialah 242.7°. 5 (a) 020° (b) 116° (c) 260° (d) 225°

(e) 318° 6 (a) 260° (b) 260°

(c) 320°


U

Q

P

230°

2 90° + 24° = 114° Bearing Q dari P ialah 114°. 3

U U

Q

P115°

65° 65°

115°

Bearing Q dari P ialah 115°.

BAB 8 Bearing

8.1 Bearing 1 (a) (i) Utara (ii) Timur (iii) Barat (b) (i) Timur laut (ii) Tenggara (iii) Selatan (iv) Barat daya (v) Barat (vi) Barat laut 2 (a) 075° (b) 140° (c) 230° (d) 220° (e) 295° 3 (a)

U A

B

70°

(b) U

A

B

110°


J17

3 (a) Beza = 88° – 38° = 50° (b) Beza = 15° 159 + 70° = 85° 159

(c) Beza = 44° 339 + 33° 449

= 77° 779 = 78° 179

9.3 Kedudukan Tempat 1 (a) P(0°, 73°T) (b) P(70°U, 47°T) (c) P(0°, 170°B) (d) P(49°S, 0°) (e) P(38° 279U, 63° 269T) 2 (a)

U

0°

16°U

S

OP

10°T

(b)

U

0°

S

OP

120°B

Latihan Bestari 9.3 1 53°U 23°T. 2

U

0°

35°U

S

O

T

76°B

3 • P = (0°, 30°T). • Q = (50°U, 70°B). • R = (40°S, 70°B).

9.4 Jarak pada Permukaan Bumi 1 (a) Panjang lengkok PQ = 26° 439 × 60 = 1 603 batu nautika (b) Panjang lengkok PQ = 67° 529 × 60 = 4 072 batu nautika 2 (a) Jarak = (80° – 10°) × 60 = 4 200 batu nautika (b) Jarak = (20° + 17° 379) × 60 = 2 257 batu nautika (c) Jarak = (17° 169 + 20° 279) × 60 = 2 263 batu nautika (d) Jarak = (86° 179 – 17° 299) × 60 = 68° 489 × 60 = 4 128 batu nautika (e) Jarak = (76° 69 – 16° 299) × 60 = 59° 379 × 60 = 3 577 batu nautika

3 (a) 1 980 ÷ 60 = 33° Q – 10° = 33° Q = 43° U (b) 1 100 ÷ 60 = 18° 209

21° 169 – Q = 18° 209

Q = 21° 169 – 18° 209

= 2° 569 U (c) 1 760 ÷ 60 = 29° 209

Q + 17° 269 = 29° 209

Q = 29° 209 – 17° 269

= 11° 549 S (d) 2 160 ÷ 60 = 36° Q = 40° 119 – 36° = 4° 119 S (e) 4 000 ÷ 60 = 66° 409

17° 209 + Q = 66° 409

Q = 66° 409 – 17° 209

= 49° 209 U 4 (a) Beza longitud = 30° 179 + 15° 299

= 45° 469

Jarak = 45° 469 × 60 = 2 746 batu nautika (b) Beza longitud = 57° 279 – 25° 179

= 32° 109

Jarak = 32° 109 × 60 = 1 930 batu nautika 5 (a) Beza longitud = 4 251 ÷ 60 = 70.85 = 70° 519

B + 2° 99 = 70° 519

B = 68° 429 B (b) Beza longitud = 5 024 ÷ 60 = 83° 449

56° 569 + B = 83° 449

B = 26° 489 T 6 (a) Beza longitud = 83° 279 – 17° 399

= 65° 489

Jarak = 65° 489 × 60 × kos 28° 109

= 3 480.5 batu nautika (b) Beza longitud = 121° 259 – 31° 179

= 90° 89

Jarak = 90° 89 × 60 × kos 63° 109

= 2 441.2 batu nautika (c) Beza longitud = 69° 179 + 17° 299

= 86° 469

Jarak = 86° 469 × 60 × kos 17° 109

= 4 974.1 batu nautika (d) Beza longitud = 120° 399 – 18° 199

= 102° 209

Jarak = 102° 209 × 60 × kos 72° 179

= 1 868.5 batu nautika

Latihan Bestari 9.1 1 140°B. 2

U

S

O

71°B

71°

3 (a) 120° – 90° = 30° (b) 6° 239 + 4° 259 = 10° 489

(c) 21° 239 –13° 419 = 7° 429

9.2 Latitud 1 (a)

0°OA

(b)

0°O

A

2 (a) 71°U (b) 66°S (c) 90°U (d) 31°U (e) 68°S

(f) 90°S (g) 0° (h) 70°U 3 (a)

0°

76°S

76°

U

S

O

(b)

0°

21° 17�U

21° 17�

U

S

O

4 (a) 81° – 59° = 22° (b) 71° 169 + 90° = 161° 169

(c) 30° + 40° = 70° (d) 41° 399 – 17° 479 = 23° 529

(e) 31° 479 + 43° 519 = 75°389

Latihan Bestari 9.2 1 (a) 70°U (b) 0° 2

0°

U

S

O

22° 9�S


J18

(e) Beza longitud = 80° – 19° 369

= 60° 249

Jarak = 60° 249 × 60 × kos 30 = 3 138.5 batu nautika 7 (a) 2 600 = θ × 60 × kos 48° 109

θ = 64° 589

B = 64° 589 + 17° 259

= 82° 239 T (b) 1500 = θ × 60 × kos 41° 299

θ = 33° 229

B + 17° 259 = 33° 229

B = 15° 579 B (c) 3 120 = θ × 60 × kos 36° 419

θ = 64° 519

81° 109 – B = 64° 519

B = 16° 199 B (d) 1 500 = θ × 60 × kos 41° θ = 33° 89

B = 33° 89 – 16° 279

= 16° 419

(e) 1 000 = θ × 60 × kos 50° θ = 25° 569

B = 25° 569 – 10° 17’ = 15° 399 T 8 (a) Jarak = (180° – 72° – 41°) × 60 = 4 020 batu nautika (b) Jarak = (180° – 72° – 65°) × 60

= 2 580 batu nautika (c) Jarak = (180° – 30° – 62°) × 60 = 5 280 batu nautika (d) Jarak = (180° – 63° – 75°) × 60 = 2 520 batu nautika (e) Jarak = (180° – 20° 169 – 78° 259) × 60 = 4 879 batu nautika 9 (a) Jarak = (78° 419 + 29° 369) × 60 = 6 497 batu nautika Masa yang diambil = 6 497 ÷ 520 = 12 jam 30 minit (b) Jarak = (30° + 72°) × 60 × kos 32° 179

= 5 174 batu nautika Masa yang diambil = 5 174 ÷ 470 = 11 jam 1 minit (c) Jarak = 410 × 4 = 1 640 batu nautika 1 640 = (62° 17’ – 29° 38’) × 60 × kos θ kos θ = 0.8372 θ = 33° 109 U (d) Jarak = (180° – 50° – 50°) × 60 = 4 800 batu nautika

Masa yang diambil = 4 800 ÷ 500 = 9 jam 36 minit (e) Jarak = 1 jam 48 minit × 520 knot = 936 batu nautika (x – 12° 259) × 60° × kos 53° = 936 (x – 12° 259) = 25° 559

x = 38° 209

Latihan Bestari 9.4 1 Jarak PQ = Beza latitud 3 60 = (61° 179 + 17° 299) 3 60 = (78° 469 3 60) = 4 726 batu nautika 2 Jarak PQ = Beza latitud 3 60 4 000 = θ 3 60 θ = 66° 409

Beza latitud = 66° 409

30° 109 + x = 66° 409

x = 66° 409 – 30° 109

= 36° 309

3 Jarak PQ = Beza longitud 3 60 3 kos (latitud) = (120° – 62° 179) 3 60 3 kos 27° 199 = [57° 4393 60)] 3 kos 27° 199

= 3 076.819 = 3 076.8 batu nautika 4 Jarak AB = Beza longitud 3 60 3 kos (latitud) 490 = θ 3 60 3 kos 41° 329

θ = 10° 559

[ Longitud B = (21° 179 + 10° 559) B = (31° 729) B= 32° 129B 5 Jarak terpendek antara P dan Q = (180° – 50° – 50°) 3 60 = 4 800 batu nautika

SUDUT KBAT 1 Daripada rajah, P = (80° 179U, 20°B) dan Q = (17° 269U, 20°B). Jarak PQ = (80° 179 – 17° 269) 3 60 = (62° 5193 60) = 3 771 batu nautika

Masa yang diambil = JarakLaju

= 3 771450

= 8.38 jam

= 8 jam 23 minit Waktu berlepas = (2210 – 0823) jam = Jam 1347

PRAKTIS BAB 9Soalan Objektif 1 C 2 B 3 B 4 B 5 C 6 C 7 D 8 A 9 A 10 B11 D 12 A 13 A 14 A 15 A16 B 17 A 18 C 19 A 20 D

Soalan Subjektif 1 (a) Longitud = (180° – 15°) B = 165°B

(b)

15°T165°B

U

42°U

42°S

S

OP

R

R ialah (42°S, 165°B). (c) Jarak terpendek = (90° – 42°) 3 60 = 2 880 batu nautika (d) (i) Jarak PM = Laju 3 masa = 480 3 6 = 2 880 batu nautika (ii) Jarak PM = θ 3 60 3 kos 42° 2 880 = θ 3 60 3 kos 42° θ = 64° 359

Longitud M = (64° 359 – 15°) B = 49° 359B 2 (a) Longitud Q = (180° – 60°) B = 120°B (b) Jarak PR = θ 3 60 5 400 = θ 3 60 θ = 90° Latitud R = (90° – 40°) U = 50°U (c) Jarak PQ = Beza longitud 3 60 3 kos 40° = 180° 3 60 3 kos 40° = 8 273.2799 = 8 273.3 batu nautika (d) Jarak Q ke P = (90° – 40°) 3 2 3 60 = 50° 3 2 3 60 = 6 000 batu nautika Jarak P ke R = 5 400 batu nautika


= 6 000 + 5 400600

= 19 jam

3 (a)

20°T160°B

U

30°S

30°U

S

OP

J

JP ialah diameter bumi, kedudukan P ialah (30°S, 160°B).

(b) (i) Jarak JL = Beza longitud 3 60 3 kos 30°

θ = 76° 599

y = 76° 599 – 20° = 56° 599

(ii) Jarak JK = Beza latitud 3 60 3 600 = a 3 60

a = 60° [ x = 60° – 30°

= 30°


J19

(c) Jarak JL = 4 000 batu nautika Jarak LM = (30° + 40°) 3 60 = 4 200 batu nautika Masa yang diambil

= JarakLaju

= 4 000 + 4 200540

= 15 jam 11 minit

4 (a) Kedudukan Q = (40°U, 30°B) Longitud R = (180° – 30°) T = 150°T Kedudukan R ialah (40°U, 150°T). (b) Jarak terpendek = Beza longitud 3 60 = (90° – 40°) 3 60 3 2 = 6 000 batu nautika (c) Jarak PX = θ 3 60 6 200 = θ 3 60 θ= 103° 209

[ Latitud X = (103° 209 – 40°) S = 63° 209S (d) Jarak PR = Beza longitud 3 60 3 kos 40° = 150° 3 60 3 kos 40° = 6 894.4 batu nautika Masa yang diambil

= JarakLaju

= 6 894.4540

= 12 jam 46 minit

5 (a) PQ ialah diameter selarian latitud. Longitud Q = (180° – 20°) B

= 160°B (b)

20°T

58°S

58°U

160°B

U

S

O

R

P

PR ialah diameter bumi. R ialah (58°S, 160°B). (c) Jarak terpendek = (90° – 58°) 3 60 = 1 920 batu nautika (d) (i) Jarak PM = Purata laju 3 Masa yang diambil = 600 3 8 = 4 800 batu nautika (ii) Jarak PM = Beza longitud 3 60 3 kos 58° 4 800 = θ 3 60 3kos 58° θ = 150° 589

[ Longitud M = (20° + 150° 589) T = 170° 589T 6 (a) Jarak PR = Beza latitud 3 60 3 600 = θ 3 60 θ= 60°

[ Latitud R = (60° – 50°) U = 10°U (b) Jarak PQ = Beza longitud 3 60 3 kos 50° = 180° 3 60 3 kos 50° = 6 942.1 batu nautika (c) PQ ialah diameter selarian latitud. Longitud Q = (180° – 40°) B = 140°B (d) Jarak QP = (90° – 50°) 3 2 3 60 = 4 800 batu nautika Jarak PR = 3 600 batu nautika Masa yang diambil

= DistanceSpeed

= 4 800 + 3 600500

= 16 jam 48 minit

7 (a)

140°T40°B

U

30°S

30°U

S

O

A

C

AC ialah diameter bumi. Maka, kedudukan C ialah (30°U, 140°T). (b) Jarak AB = (70° – 40°) 3 60 3 kos 30° = 1 558.8 batu nautika (c) Jarak AQ = θ 3 60 3 kos 30° 5 000 = θ 3 60 3 kos 30° θ= 96° 139

[ Longitud Q = (96° 139 – 40°) T = 56° 139T 8 (a) (i) Longitud R = (180° – 30°) B = 150°B (ii) Jarak PQ = (40° + 30°) 3 60 3 kos 21° = 3 921 batu nautika

(b) Jarak QX = θ 3 60 4 800 = θ 3 60 θ= 80° [ Latitud X = (80° – 21°) U = 59°U 9 (a) (i) 30°S. (ii) Longitud P = (180° – 24°) T = 156°T (b) Jarak PQ = 180° 3 60 3 kos 30° = 9 353.1 batu nautika (c) Jarak terpendek PQ = (180° – 30° – 30°) 3 60 = 7 200 batu nautika


= 7 200450

= 16 jam

Masa = 0230 + 1600 = 1830 jam 10 (a) Latitud Q ialah 35°S. (b) Jarak = (90° – 35°) 3 60 = 3 300 batu nautika (c) Jarak PX = θ 3 60 3 000 = θ 3 60 θ= 50° [ Latitud X = (50° – 35°) S = 15°S (d) Jarak PY = (70° – 42°) 3 60 3 kos 35° = 1 376.2 batu nautika


= 1 376.2320

= 4 jam 18 minit

BAB 10 Pelan dan Dongakan

10.1 Unjuran Ortogon 1 (a)

QP

RS

(b)

QP

RS

(c)

QP

RS

(d)

QP

RS

(e)

PQ

RS


J20


QP

RS

2

C

A B

E

D

PQ

SR

10.2 Pelan dan Dongakan 1 (a)

F/BE/A 5 cm

G/CH/D

3 cm

(b)

E

A

F/B

H

D

G/C

3 cm

2 cm

6 cm

(c)

L/G/B

M/H/C

E/A

J/D

4 cm

K/F

N/I

4 cm2 cm

(d)

H/B

I/C

E/A

L/D

4 cm

F G

K J

2 cm3 cm 1 cm (e)

G/B

H/C

E/A

J/D

4 cm

F

I

2 cm 3 cm

2 (a) (i)

B/C

V

5 cm

6 cm

A/D

(ii)

6 cm

C/D

V

4 cmB/A

(b) (i)

B/C

G/H

L/M

A/D

E/J

K/N

F/I

3 cm

3 cm

4 cm

6 cm

(ii)

C/D

M/N

H/I/J

B/A

L/K

G/F/E

3 cm

4 cm

3 cm

(c) (i)

B/C

V

5 cm

4 cm

A/D

(ii)

C/D

V

4 cm

4 cm

B/A

(d) (i)

B/C

H/IG/J

6 cmA/D

F/K3 cm

1 cm

2 cm3 cm

E/L

(ii)

A/B4 cmD/C

E/F

G/HJ/I

L/K

2 cm3 cm

1 cm

(e) (i)

B/C

G/H

3 cm

A/D

E/JF/I2 cm

4 cm

5 cm

(ii)

A/B4 cmD/C

E/F

G1 cm

3 cm

H

J/I

3 (a)

Q10 cm

7.5 cm

P

RT/S

(b)

B/C14 cmA/D

F/G4 cm

15 cm

E/H

(c)

M/N

U

10 cm

2 cm

5 cm

S/T

L/K

P

R/Q

SUDUT KBAT 1 (a)

B

E

F

DC

5 cm

6 cm 10 cm

A

(b)

B

E F

J I

H

C

G

D

6 cm

5 cm

3 cm

2 cm

10 cm

A

Latihan Bestari 10.2

1 (a)

F/E

G/H

L/B

M/C

K/A

N/D

3 cm 3 cm

4 cm


J21

(b)

E/H

F/G

3 cm 3 cm

2 cm

B/C

L/M

3 cm

A/D

K/N

5 cm

(c)

A/E/B

F

L

K

D/H/C

G

M

N

2 cm

4 cm

2 cm

1 cm

PRAKTIS BAB 10Soalan Subjektif 1 (a)

E/L

F/K

3 cm

4 cm

2 cmI/A

J/B

H/D

G/C

(b) (i)

A/B/P

I/JL/K

E/F

M/N

D/C/Q

H/G/R

4 cm

2 cm

2 cm

2 cm

5 cm

6 cm

(ii)

P/Q

N/RM/G

B/C4 cm 2 cmA/D

I/L

H

E

J/K

F

6 cm

2 cm

2 cm

2 cm

2 (a)

G/HE/JF/I

A/BD/C 5 cm

2 cm2 cm

1.5 cm3 cm

(b) (i)

G/AE/D

J/C IH/B P/K

S/NR/M Q/L

F 3 cm

3 cm2 cm

2 cm2 cm

2cm

(ii)

L/MA/D

G/F/E

H/S/I/J

K/B/N/C

R

QP

2 cm

2 cm3 cm

4 cm

3 (a)

B/C

G/H

J/I

A/D

F/E

6 cm

2 cm

4 cm4 cm

(b) (i)

K/A

Q/M

P/N

G/L

J/B

E/D

H

I/C2 cm

4 cm

2 cm3 cm

(ii)

A/BD/C 5 cm

1 cm

2 cmK/L

P/Q

N/M

E/H

2 cm

2 cm

G

JI

4 (a)

B/C

F

G

A/D

E

H2 cm

4 cm

5 cm

(b) (i)

C/D

G/H

6 cm

1 cm

J/B/A

F/E

4 cm4 cm

5 cm L/K

M/N

(ii)

J

M/LN/K

5 cm

E/A

H/D G/C

6 cm

1 cm

5 cm F/B

5 (a)

5 cm

4 cm2 cm

B/C

I/JE/H

F/G

A/D

(b) (i)

N/K

M/L

E/A

H/DG Q/J/C

F

I/P/B

2.5cm

2.5cm

2cm

6 cm

(ii)

2 cm

2 cm

2 cm

G

J/H2 cm

1 cm FN/P

I/E

M/Q

1 cm6 cm L/C/DK/B/A

6 (a)

4 cm

5 cm B/CA/D2 cm

2 cmH

F1 cm

G

E

(b) (i)

H/BS/E/A 5 cm

2cm

F/C

6 cm

R/M

Q/L P/N/K

G/D

4 cm

(ii)

C/D

K/L

P/Q

FN

G

4 cm

4 cm

4 cm

2 cmB/A/M

H

E

S/R

2 cm

2 cm

1 cm1 cm

7 (a)

B/C

G

F

2 cm

4 cmA/D

H

E

2 cm

3 cm

(b) (i)

E/A

H/D

F/B

K/G/C

L/M/N

6 cm

3 cm

4 cm

1cm

(ii)

B/A

F/E

5 cm

G/M/H

C/N/D

K/L

6 cm

2 cm

6 cm1.5cm

8 (a)

B/C

E

F

3 cm

2 cm

5 cm

3 cm

3 cm3 cm

3 cm

A/D

I

J

G

H

3 cm

2 cm

(b) (i)

E/B

F/C

I/A

J/D

6 cm

2.5cm

2.5cm

H


J22

(ii)

C/D

F/JH

E/I

B/A 6 cm

5 cm3 cm

Kertas Model SPM

KERTAS 1 1 A 2 D 3 A 4 A 5 B 6 A 7 C 8 D 9 D 10 B11 B 12 B 13 D 14 C 15 C16 A 17 B 18 A 19 B 20 C21 D 22 B 23 B 24 A 25 B26 A 27 B 28 B 29 D 30 C31 C 32 A 33 A 34 D 35 A36 B 37 A 38 B 39 C 40 D

KERTAS 2 Bahagian A

1 p = 3p + 2

p(p + 2) = 3 p2 + 2p = 3 p2 + 2p – 3 = 0 (p + 3)(p – 1) = 0 p = –3 atau p = 1 2 4x – 3y = 5 … 1

2x – y = 2 … 2 2 3 2 : 4x – 2y = 4 … 3 3 – 1 : y = –1 Gantikan y = –1 ke dalam 2 . 2x – (–1) = 2 2x = 2 – 1

x = 12

[ x = 12

dan y = –1

3

x

y = 2x

y = 6

y

0

6

4

D C

H G6 cm

5 cm

�

tan = 56

[ = 39° 489

DHC = 90° – = 90° – 39° 489

= 50° 129

5 (a) –x + 2y = 6 Apabila y = 0, –x = 6 x = –6 Maka, persamaan TU ialah x = –6.

(b) –x + 2y = 6 2y = x + 6

y = 12

x + 3

[ Kecerunan PQ = Kecerunan RS

= 12

y – (–3) = 12

[x – (–4)]

y + 3 = 12

(x + 4)

y = 12

x + 2 – 3

y = 12

x – 1

6 Isi padu kon

= 13

3 227

3 32 3 4

= 2647

cm3

Isi padu silinder

= 227

3 72 3 10 cm

= 1 540 cm3

[ Isi padu baki pepejal

= 1 540 – 2647

= 1 502 2

7 cm3

7 (a) Panjang lengkok RQP

= 110°360°

3 2(3.14)(8)

= 15.35 cm [ Perimeter seluruh rajah = 15.35 cm + 8 cm + 8 cm = 31.35 cm (b) Luas sektor ORQP

= 110°360°

3 (3.14)(8)2

= 61.4 cm2

Luas sektor OTVW

= (180° – 75°)360°

3 (3.14)(4)2

= 105°360°

3 (3.14)(16)

= 14.65 cm2

[ Luas kawasan berlorek = 61.4 – 14.65 = 46.75 cm2

8 (a) (i) 3 3 4 = 12 (Benar) 34 = 64 (Tidak benar kerana 34 = 81) Maka, pernyataan

3 3 4 = 12 dan 34 = 64 ialah palsu. (ii) {2} {1, 2, 3} (Benar) {2} {1, 2, 3} = {1, 2, 3} (Palsu kerana {2} {1, 2, 3} = {2}) Maka, pernyataan itu benar. (b) Premis 2 ialah: 2x – 3 . 7 (c) Jika x = 8, maka 2x = 16. Jika 2x = 16, maka x = 8.

9 (a) P(Alvin memilih Sekolah P) = 34

P(Bakar memilih Sekolah P) = 1 – 23

= 13

[ P(Alvin dan Bakar memilih Sekolah P)

= 34

3 13

= 14

(b) P(Alvin memilih Sekolah Q)

= 1 – 34

= 1

4 [ P(Alvin memilih Sekolah Q atau

Bakar memilih Sekolah P)

= 14

+ 13

= 3 + 4

12 = 7

12 10 (a) 4 s. (b) 108 = (t – 4) 3 18 108 = 18t – 72 18t = 180 t = 10 (c) Kadar perubahan = 0 – 18

12 – 10

= –18

2

= –9 m s–2

11 (a) AB = 1 00 1 , maka B = A–1.

B = 1(–1)(1) – (3)(2)

1 –2–3 –1

= 1–1 – 6

1 –2–3 –1

= – 17

1 –2–3 –1

Maka, k = 7 dan h = –1.

(b) –1 2 3 1 x

y = –4 5

Axy = –4

5 A–1Ax

y = A–1–4 5

xy = – 1

7 1 –2–3 –1 –4

5 = – 1

7 –14 7

= 2–1

Maka, x = 2 dan y = –1.

Bahagian B 12 (a) y = 2x2 – 3x – 4 Apabila x = –1, y = 2(–1)2 – 3(–1) – 4 = 2(1) + 3 – 4 = 1


J23

Apabila x = –0.5, y = 2(–0.5)2 – 3(–0.5) – 4 = 2(0.25) + 1.5 – 4 = –2 Apabila x = 2, y = 2(2)2 – 3(2) – 4 = 8 – 6 – 4 = –2

x –2 –1 –0.5 0 1 2 3 4

y 10 1 –2 –4 –5 –2 5 16

(b)

y

x2 3 4–1 0 1–2

16

14

10

8

6

4

2

–2

–4

–6

12

–0.65 1.5–1.35

2.4

2.9

3.8

y = 12

x – 1

(c) (i) Daripada graf, apabila x = 1.5,

y = –3.7. (ii) Daripada graf, apabila

y = 3.8, x = –1.35 dan 2.9. (d) 4x2 – 7x – 6 = 0 … 1 y = 2x2 – 3x – 4 … 2 1 4 2 : 2x2 – 7

2x – 3 = 0 … 3

3 – 2 : 2x2 – 72

x – 3 = 0

(–) 2x2 – 3x – 4 = y

– 12

x + 1 = –y

y = 12

x – 1

Lukis graf y = 12

x – 1.

x 0 2

y –1 0

Nilai yang memuaskan 4x2 – 7x – 6 = 0 ialah –0.65 dan 2.4.

13 (a) (i)

6

y

xP�(5, 0)

P4

2

2

4

–2 0–4 2 4 6

90°(2, 2)

(ii) (a)

6

y

x

Q

Q�(–3, 2)

Q�

4

2

2

4

–2 0–4 2 4 6

(b)

6

y

x

Q

Q�(–1, 5)

Q�

4

2 (2, 2)

90°

2

4

–2 0–4 2 4 6

P

(b) (i) (a)

6

y

xy = 1B

AC

DF

E

4

2

2

4

6

0 2 4 6 8

Penjelmaan V ialah pantulan pada garis y = 1. (b)

6

y

x

D HF

G

I

E

4

2

2

4

6

0 2 4 6 8

(4, 1)

Penjelmaan W ialah pembesaran dengan faktor skala 2 pada pusat

(4, 1).

(b) (ii) Anggap luas ABC sebagai x.

34.24x

= 22

x = 34.244

= 8.56 Maka, luas ABC ialah 8.56 cm2. 14 (a) (i)

Skor, x Kekerapan, f Kekerapan

fx longgokan

0 8 8 0

1 15 23 15

2 24 47 48 3 12 59 36

4 9 68 36

5 4 72 20

Median = nilai ke- 722

= nilai ke-36 = 2 (ii) Min

= 15 + 48 + 36 + 36 + 2072

= 2.152777 = 2.153

(b) (i)

Markah Kekerapan

Kekerapan Sempadan longgokan atas

0 – 9 2 2 9.5

10 – 19 9 11 19.5

20 –29 15 26 29.5

30 – 39 10 36 39.5

40 – 49 8 44 49.5

50 – 59 6 50 59.5

(ii)

50

45

35

30

25

0

5

40

12.5

20

15

10

Kekerapan longgokan

37.5

19.5 29.5 39.59.5 49.5 59.525 35

Q1 Q3

Markah

(a) Q1 = nilai ke- 504

= nilai ke-12.5 = 21


J24

Q3 = nilai ke- 34

3 50 = nilai ke-37.5 = 41.5

Julat antara kuartil = Q3 – Q1

= 41.5 – 21 = 20.5

(b) = 32 – 18 = 14

15 (a)

C/D

J/I

4 cm

B/A

H/G

K/T

E/F3 cm 3 cm

2 cm

5 cm

(b) (i)

J/C

I/D

4 cm

P/B

N/A

K/EL

T/FM

1 cm 3 cm2 cm

(ii)

B/E/C4 cmA/F/D

JI

N/M

T

P/L

K1 cm

2 cm

1 cm

16 (a) (i) R = (62°U, 148°T) (ii) Q = (62°S, 148°T)

(b) Jarak PT = 2 850 batu nautika 2 850 = (Beza longitud) 3 60 3 kos 62°

Beza longitud = 101° 119

Longitud T = (101° 119 – 32°) T = 69° 119T

(c) Jarak terpendek antara P dan R = (180° – 62° – 62°) 3 60

= 3 360 batu nautika

(d) Masa

= 3 360 batu nautika570 knots

= 5 jam 54 minit

Masa kapal terbang tiba di R = 0700 + 0554 = Jam 1254

Soalan KBAT

Bab 1 (a) 13565 = 1 × 5n

n = 3(b) 13565 = (1 × 53) + (3 × 52) + (5 × 51) + (6 × 50) = 23110 23110 = 111001112

Bab 2m = 3, n = –9

Bab 3

θ = sin–135

= 36.87Sudut putaran = 36.87 × 3 = 110.61°

Bab 42D + 3R = 273D + 4R = 38

2 33 4 D

R = 2738

DR = 2 3

3 4 –1

2738

= 1(2)(4) – (3)(3) 4 –3

–3 2 2738

= –1 4(27) + (–3)(38)(–3)(27) + (2)(38)

= –1 –6–5

D = 6 R = 5

Bab 8Bearing L dari K= 360° – 125° – 60°

U

K

L

J55°

60°

125°= 175°

5

Oθ

V

W

3


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