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J1 Global Mediastreet Sdn. Bhd. (762284-U) BAB 1 Bentuk Piawai 1.1 Angka Bererti 1 (a) 5 angka bererti (b) 3 angka bererti (c) 2 angka bererti (d) 5 angka bererti (e) 3 angka bererti 2 (a) 159 000 (b) 46 000 (c) 0.46 (d) 20.3 (e) 0.00059 3 Bundarkan kepada 3 angka bererti Faktor penghubung 4.10376 501 654 as (a) 4.10 (b) 502 000 as 87 632 0.001367 (c) 87 600 (d) 0.00137 as 4 (a) 0.347 ÷ 100 × 6.8 = 0.00347 × 6.8 = 0.023596 = 0.0236 [3 angka bererti] (b) 7.2 × 12.3 ÷ 4 = 88.56 ÷ 4 = 22.14 = 22.1 [3 angka bererti] (c) 4 207 × 8 – 432 = (4 207 × 8) – 432 = 33 656 – 432 = 33 224 = 33 200 (d) 4.32 × 0.08 + 5.216 = 0.3456 + 5.216 = 5.5616 = 5.56 [3 angka bererti] (e) 2.34 × 4.11 ÷ 6 = 9.6174 ÷ 6 = 1.6029 = 1.60 [3 angka bererti] (f) 65.317 ÷ 6.2 ÷ 0.04 = 10.535 ÷ 0.04 = 263.375 = 260 [2 angka bererti] (g) 234 × 25 + 678 = 5 850 + 678 = 6 528 = 6 500 [2 angka bererti] 5 (a) 512 ÷ 16 × 8 = 32 × 8 = 256 = 260 [2 angka bererti] (b) 2 231 – 1 034 + (4 × 212) = 1 197 + 848 = 2 045 = 2 050 [3 angka bererti] Latihan Bestari 1.1 1 2 340 × 12 ÷ 6 = 28 080 ÷ 6 = 4 680 = 4 700 [2 angka bererti] 2 54 662 – 540 ÷ 9 = 54 662 – 60 = 54 602 = 55 600 [2 angka bererti] 3 (23.4 × 2.7) =7.95 [3 angka bererti] 4 (31 872 × 64) 16 = 2 039 808 ÷ 16 = 127 488 = 127 000 [3 angka bererti] 1.2 Bentuk Piawai 1 (a) 5.64 3 10 –2 (b) 2.187 3 10 –3 (c) 7.658 3 10 –4 (d) 4.5 3 10 5 (e) 8.679 3 10 4 2 (a) 61 200 (b) 0.02135 (c) 5 678 (d) 0.0007324 (e) 0.005213 3 (a) 2.3 × 10 –2 + 5.6 × 10 –3 = 2.3 × 10 –2 + 0.56 × 10 –2 = 2.86 × 10 –2 (b) 7 × 10 4 – 2.3 × 10 4 = 4.7 × 10 4 (c) 5.9 × 10 3 + 4.1 × 10 2 = 5.9 × 10 3 + 0.41 × 10 3 = 6.31 × 10 3 (d) 6.3 × 10 –3 – 9.1 × 10 –4 = 6.3 × 10 –3 – 0.91 × 10 –3 = 5.39 × 10 –3 (e) 2.5 × 10 –2 – 3.6 × 10 –3 = 2.5 × 10 –2 – 0.36 × 10 –2 = 2.14 × 10 –2 (f) 2.1 × 10 6 + 3.8 × 10 5 = 2.1 × 10 6 + 0.38 × 10 6 = 2.48 × 10 6 (g) 4.7 × 10 5 – 1.8 × 10 4 = 4.7 × 10 5 – 0.18 × 10 5 = 4.52 × 10 5 (h) 3.5 × 10 3 + 420 = 3.5 × 10 3 + 4.2 × 10 2 = 3.5 × 10 3 + 0.42 × 10 3 = 3.92 × 10 3 4 (a) 6.2 × 10 –5 × 2.8 × 10 –3 = 1.736 × 10 –7 (b) 9 × 10 4 × 1.6 × 10 3 = 14.4 × 10 7 = 1.44 × 10 8 (c) 2.64 × 10 –2 ÷ 4 × 10 –3 = 0.66 × 10 1 = 6.6 (d) 7.2 × 10 4 ÷ 8 × 10 –4 = 0.9 × 10 8 = 9 × 10 7 5 (a) Lebar buku = 4.6 × 10 3 ÷ 230 = 4.6 × 10 3 ÷ 2.3 × 10 2 = 2 × 10 1 mm (b) 5.6 × 10 –4 kg ÷ 245 980 = 2.277 × 10 –9 kg Latihan Bestari 1.2 1 (a) 2.34 3 10 5 (b) 5.21 3 10 –4 (c) 6.501 3 10 6 (d) 2.1 3 10 –6 2 (a) 51 600 (b) 0.0000602 (c) 3 000 000 (d) 0.008912 3 (a) 70 000 + 56 400 = 7 3 10 4 + 5.64 3 10 4 = (7 + 5.64) 3 10 4 = 12.64 3 10 4 = 1.264 3 10 1 3 10 4 = 1.264 3 10 5 (b) 7.6 3 10 4 – 2.3 3 10 3 = 7.6 3 10 4 – 0.23 3 10 4 = (7.6 – 0.23) 3 10 4 = 7.37 3 10 4 (c) 0.065 3 0.3 = 6.5 3 10 –2 3 3 3 10 –1 = (6.5 3 3) 3 10 –2 + (–1) = 19.5 3 10 –3 = 1.95 3 10 1 3 10 –3 = 1.95 3 10 –2 (d) 5.04 3 10 5 4 7 3 10 –2 = (5.04 4 7) 3 10 5 – (–2) = 0.72 3 10 7 = 7.2 3 10 –1 3 10 7 = 7.2 3 10 6 SUDUT KBAT Isi padu setiap kubus = 343 cm 3 ÷ 8 = 42.875 cm 3 Maka, panjang tepi kubus = 3 42.875 = 3.5 cm Luas setiap muka kubus = 3.5 2 cm 2 = 12.25 cm 2 Maka, jumlah luas permukaan bongkah kayu dalam Rajah Q = 12.25 cm 2 3 24 = 294 cm 2 = 2.94 3 10 2 cm 2 PRAKTIS BAB 1 Soalan Objektif 1 A 2 B 3 B 4 A 5 B 6 D 7 D 8 D 9 D 10 B 11 C 12 A 13 A 14 A 15 D 16 B 17 D 18 A 19 B 20 D BAB 2 Ungkapan dan Persamaan Kuadratik 2.1 Ungkapan Kuadratik 1 2k 2 + 3, 6x + 4x 2 , t 2 + 4 – 8t, y 2 + 2y 2 (a) (5y + 4) 1 5 y = y 2 + 4 5 y (b) 3(w + 2) 2 = 3(w 2 + 4w + 4) = 3w 2 + 12w + 12 (c) (–2y – 3) 2 = 4y 2 + 12y + 9 (d) (2y – 1)(2y – 2) = 4y 2 – 4y – 2y + 2 = 4y 2 – 6y + 2 (e) (4p + 9)(p – 1) = 4p 2 – 4p + 9p – 9 = 4p 2 + 5p – 9 3 (a) Luas segi tiga = 1 2 × (2x + 1)(x – 3) = 1 2 × (2x 2 – 6x + x – 3) = 1 2 × (2x 2 – 5x – 3) = x 2 5 2 x 3 2 cm 2 (b) Luas segi empat sama = (q – 2)(q – 2) = (q 2 – 4q + 4) cm 2 Latihan Bestari 2.1 1 (a) Ya (b) Ya (c) Bukan (d) Ya 2 (a) 6x(x – 5) = 6x 2 – 30x (b) –5x(x + 2) = –5x 2 – 10x (c) (3x – 4)(3x – 4) = 9x 2 – 12x – 12x + 16 = 9x 2 – 24x + 16 (d) (2x + 3)(3x – 4) = 6x 2 – 8x + 9x – 12 = 6x 2 + x – 12 3 (x – 3)(4x + 1) = 4x 2 + x – 12x – 3 = 4x 2 – 11x – 3 JAWAPAN Anjakan Prima Math F4 Jaw 4th.indd 1 9/10/2017 3:50:28 PM

Transcript of JAWAPAN - mediastreet.com.mymediastreet.com.my/impak_A+_jawapan/impak A+ Maths Tg 4 -Jawapan.pdf ·...

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BAB 1 Bentuk Piawai

1.1 Angka Bererti 1 (a) 5 angka bererti (b) 3 angka bererti (c) 2 angka bererti (d) 5 angka bererti (e) 3 angka bererti 2 (a) 159 000 (b) 46 000 (c) 0.46 (d) 20.3 (e) 0.00059 3 Bundarkan

kepada 3 angka berertiFaktor

penghubung

4.10376 501 654as

(a) 4.10 (b) 502 000

as

87 632 0.001367

(c) 87 600 (d) 0.00137

as

4 (a) 0.347 ÷ 100 × 6.8 = 0.00347 × 6.8 = 0.023596 = 0.0236 [3 angka bererti] (b) 7.2 × 12.3 ÷ 4 = 88.56 ÷ 4 = 22.14 = 22.1 [3 angka bererti] (c) 4 207 × 8 – 432 = (4 207 × 8) – 432 = 33 656 – 432 = 33 224 = 33 200 (d) 4.32 × 0.08 + 5.216 = 0.3456 + 5.216 = 5.5616 = 5.56 [3 angka bererti] (e) 2.34 × 4.11 ÷ 6 = 9.6174 ÷ 6 = 1.6029 = 1.60 [3 angka bererti] (f) 65.317 ÷ 6.2 ÷ 0.04 = 10.535 ÷ 0.04 = 263.375 = 260 [2 angka bererti] (g) 234 × 25 + 678 = 5 850 + 678 = 6 528 = 6 500 [2 angka bererti] 5 (a) 512 ÷ 16 × 8 = 32 × 8 = 256 = 260 [2 angka bererti] (b) 2 231 – 1 034 + (4 × 212) = 1 197 + 848 = 2 045 = 2 050 [3 angka bererti]

Latihan Bestari 1.1 1 2 340 × 12 ÷ 6 = 28 080 ÷ 6 = 4 680 = 4 700 [2 angka bererti] 2 54 662 – 540 ÷ 9 = 54 662 – 60 = 54 602 = 55 600 [2 angka bererti] 3 √(23.4 × 2.7) =7.95 [3 angka bererti] 4 (31 872 × 64)

16 = 2 039 808 ÷ 16

= 127 488 = 127 000 [3 angka bererti]

1.2 Bentuk Piawai 1 (a) 5.64 3 10–2 (b) 2.187 3 10–3

(c) 7.658 3 10–4 (d) 4.5 3 105

(e) 8.679 3 104 2 (a) 61 200 (b) 0.02135 (c) 5 678 (d) 0.0007324 (e) 0.005213 3 (a) 2.3 × 10–2 + 5.6 × 10–3

= 2.3 × 10–2 + 0.56 × 10–2

= 2.86 × 10–2

(b) 7 × 104 – 2.3 × 104 = 4.7 × 104

(c) 5.9 × 103 + 4.1 × 102

= 5.9 × 103 + 0.41 × 103

= 6.31 × 103

(d) 6.3 × 10–3 – 9.1 × 10–4

= 6.3 × 10–3 – 0.91 × 10–3

= 5.39 × 10–3

(e) 2.5 × 10–2 – 3.6 × 10–3

= 2.5 × 10–2 – 0.36 × 10–2

= 2.14 × 10–2

(f) 2.1 × 106 + 3.8 × 105

= 2.1 × 106 + 0.38 × 106

= 2.48 × 106

(g) 4.7 × 105 – 1.8 × 104

= 4.7 × 105 – 0.18 × 105

= 4.52 × 105

(h) 3.5 × 103 + 420 = 3.5 × 103 + 4.2 × 102

= 3.5 × 103 + 0.42 × 103

= 3.92 × 103

4 (a) 6.2 × 10–5 × 2.8 × 10–3 = 1.736 × 10–7

(b) 9 × 104 × 1.6 × 103 = 14.4 × 107

= 1.44 × 108

(c) 2.64 × 10–2 ÷ 4 × 10–3 = 0.66 × 101

= 6.6 (d) 7.2 × 104 ÷ 8 × 10–4 = 0.9 × 108

= 9 × 107

5 (a) Lebar buku = 4.6 × 103 ÷ 230 = 4.6 × 103 ÷ 2.3 × 102

= 2 × 101 mm (b) 5.6 × 10–4 kg ÷ 245 980 = 2.277 × 10–9 kg

Latihan Bestari 1.2 1 (a) 2.34 3 105 (b) 5.21 3 10–4

(c) 6.501 3 106 (d) 2.1 3 10–6

2 (a) 51 600 (b) 0.0000602 (c) 3 000 000 (d) 0.008912 3 (a) 70 000 + 56 400 = 7 3 104 + 5.64 3 104

= (7 + 5.64) 3 104

= 12.64 3 104

= 1.264 3 101 3 104

= 1.264 3 105

(b) 7.6 3 104 – 2.3 3 103

= 7.6 3 104 – 0.23 3 104

= (7.6 – 0.23) 3 104

= 7.37 3 104

(c) 0.065 3 0.3 = 6.5 3 10–2 3 3 3 10–1

= (6.5 3 3) 3 10–2 + (–1)

= 19.5 3 10–3

= 1.95 3 101 3 10–3

= 1.95 3 10–2

(d) 5.04 3 105 4 7 3 10–2

= (5.04 4 7) 3 105 – (–2)

= 0.72 3 107

= 7.2 3 10–1 3 107

= 7.2 3 106

SUDUT KBATIsi padu setiap kubus = 343 cm3 ÷ 8 = 42.875 cm3

Maka, panjang tepi kubus = 3√42.875 = 3.5 cmLuas setiap muka kubus = 3.52 cm2 = 12.25 cm2

Maka, jumlah luas permukaan bongkah kayu dalam Rajah Q = 12.25 cm2 3 24 = 294 cm2

= 2.94 3 102 cm2

PRAKTIS BAB 1Soalan Objektif 1 A 2 B 3 B 4 A 5 B 6 D 7 D 8 D 9 D 10 B11 C 12 A 13 A 14 A 15 D16 B 17 D 18 A 19 B 20 D

BAB 2Ungkapan dan Persamaan Kuadratik

2.1 Ungkapan Kuadratik 1 2k2 + 3, 6x + 4x2, t2 + 4 – 8t, y2 + 2y

2 (a) (5y + 4) 15

y = y2 + 45

y (b) 3(w + 2)2 = 3(w2 + 4w + 4) = 3w2 + 12w + 12 (c) (–2y – 3)2 = 4y2 + 12y + 9 (d) (2y – 1)(2y – 2) = 4y2 – 4y – 2y + 2 = 4y2 – 6y + 2 (e) (4p + 9)(p – 1) = 4p2 – 4p + 9p – 9 = 4p2 + 5p – 9

3 (a) Luas segi tiga = 12

× (2x + 1)(x – 3) = 1

2 × (2x2 – 6x + x – 3)

= 1

2 × (2x2 – 5x – 3)

= x2 – 5

2 x – 3

2 cm2

(b) Luas segi empat sama = (q – 2)(q – 2) = (q2 – 4q + 4) cm2

Latihan Bestari 2.1 1 (a) Ya (b) Ya (c) Bukan (d) Ya 2 (a) 6x(x – 5) = 6x2 – 30x (b) –5x(x + 2) = –5x2 – 10x (c) (3x – 4)(3x – 4) = 9x2 – 12x – 12x + 16 = 9x2 – 24x + 16 (d) (2x + 3)(3x – 4) = 6x2 – 8x + 9x – 12 = 6x2 + x – 12 3 (x – 3)(4x + 1) = 4x2 + x – 12x – 3 = 4x2 – 11x – 3

JAWAPAN

Anjakan Prima Math F4 Jaw 4th.indd 1 9/10/2017 3:50:28 PM

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2.2 Pemfaktoran Ungkapan Kuadratik 1 (a) 2(y2 – 6) (b) 8(2 – t2) (c) –2(3 + y2) (d) 7m(1 – 3m) (e) y(4y – 5) (f) 3u(–4u + 1) 2 (a) (1 – x)(1 + x) (b) (3 – x)(3 + x) (c) (x – 7)(x + 7)

(d) 1 – 12 x 1 +

12 x

(e) 2(4 – x)(4 + x) 3 (a) (x + 6)(x – 1) (b) (x + 2)(x – 3) (c) (x + 3)(x + 7) 4 (a) 4x2 + 12x + 8 = 4(x2 + 3x + 2) = 4(x + 1)(x + 2) (b) 2x2 – 6x – 8 = 2(x2 – 3x – 4) = 2(x – 4)(x + 1) (c) 6x2 – 2x – 4 = 2(3x2 – x – 2) = 2(3x + 2)(x – 1) (d) 3x2 + 18x + 24 = 3(x2 + 6x + 8) = 3(x + 2)(x + 4) (e) 2x2 – 2x – 4 = 2(x2 – x – 2) = 2(x + 1)(x – 2)

Latihan Bestari 2.2 1 (a) –3 + 6x2 = –3(1 – 2x2) (b) 5x2 + 25 = 5(x2 + 5) (c) –6 – 36x2 = –6(1 + 6x2) (d) x2 + 8x = x(x + 8) (e) –16x2 + 8x = 8x(–2x + 1) 2 (a) –16 + x2 = x2 – 16 = (x – 4)(x + 4) (b) 9x2 – 1 = (3x – 1)(3x + 1) (c) 3x2 – 12 = 3(x2 – 4) = 3(x – 2)(x + 2) 3 (a) x2 + 4x – 12 = (x + 6)(x – 2) (b) x2 – 8x + 16 = (x – 4)(x – 4) (c) –6x + x2 + 9 = x2 – 6x + 9 = (x – 3)(x – 3) 4 (a) 3x2 + 4x – 4 = (3x – 2)(x + 2) (b) –6x + 3x2 – 9 = 3x2 – 6x – 9 = 3(x2 – 2x – 3) = 3(x – 3)(x + 1) (c) 2x2 – 6x + 4 = 2(x2 – 3x + 2) = 2(x – 1)(x – 2)

2.3 Persamaan Kuadratik 1 (a) Bukan (b) Ya (c) Ya (d) Bukan (e) Bukan 2 (a) 4x2 + x – 8 = 0 (b) 9x2 + 5x – 6 = 0 (c) 4x2 – x + 1 = 0 (d) 14x2 + 21x + 12 = 0 (e) 30x2 – 61x + 28 = 0 3 (a) x2 – 3x – 10 = 0 (b) 22r2 – 35 = 0

Latihan Bestari 2.3 1 (a) Ya (b) Bukan (c) Ya 2 (a) (4x – 1)(x + 9) = 9 4x2 + 36x – x – 9 = 9 4x2 + 35x – 18 = 0

(b)

1y

+

1y – 1

= 1

y – 1 + yy(y – 1)

= 1

2y – 1 = y2 – y y2 – y – 2y + 1 = 0 y2 – 3y + 1 = 0 3 x 3 x = 625 x2 = 625 x2 – 625 = 0 4 Anggap lebarnya sebagai x, panjang = x + 5 Luas segi empat tepat = panjang 3 lebar 40 = (x + 5) 3 x 40 = x2 + 5x x2 + 5x – 40 = 0

2.4 Punca-punca Persamaan Kuadratik 1 (a) Ya (b) Tidak 2 (a) 8x2 – 2x – 10 = 0 2(4x2 – x – 5) = 0 2(4x – 5)(x + 1) = 0

x = –1 @ 54

(b) 2x2 – 3x + 1 = 0 (2x – 1)(x – 1) = 0

x = 12

@ 1

(c) x2 – 4x + 4 = 0 (x – 2)(x – 2) = 0 x = 2 3 (a) 19.14 (b) (x – 3) × x = 28 x2– 3x = 28 x2 – 3x – 28 = 0 (x – 7)(x + 4) = 0 x = 7

Latihan Bestari 2.4 1 Gantikan x = 3 ke sebelah kiri persamaan 2(3)2 – 5(3) – 7 = 2(9) – 15 – 7 = 18 – 15 – 7 = –4 Oleh kerana kiri kanan, x = 3 bukan punca

bagi persamaan itu.

2 Gantikan y = 13 ke sebelah kiri persamaan

(1 + 3y)(1 + 3y) = 1 + 6y + 9y2

= 1 + 6 1

3 + 9 13

2

= 1 + 2 + 9 1

9 = 1 + 2 + 1 = 4

Oleh kerana kiri kanan, y = 13 bukan

punca persamaan itu.

3 Faktor bagi 9 ialah 1, 3 dan 9.

x y2 + 6y + 9

–3 (–3)2 + 6(–3) + 9 = 0

–1 (–1)2 + 6(–1) + 9 = 4

1 (1)2 + 6(1) + 9 = 16

3 (3)2 + 6(3) + 9 = 36

Dengan itu, –3 adalah punca bagi x2 + 6x + 9. 4 (a) 2x2 – 7x + 5 = 0 (2x – 5)(x – 1) = 0

x = 52

@ 1

(b) 3x2 – 6x – 9 = 0 3(x2 – 2x – 3) = 0 3(x – 3)(x + 1) = 0 x = –1 @ 3 (c) 12x2 – 9x – 21 = 0 3(4x2 – 3x – 7) = 0 3(4x – 7)(x + 1) = 0

x = 74

@ –1

5 Januari = x Februari = x + 5 x(x + 5) = 150 x2 + 5x – 150 = 0 (x + 15)(x – 10) = 0 x = –15 @ 10 x = –15 tidak boleh menjadi jawapan kerana

mempunyai nilai negatif. Maka, nilai yang mungkin bagi x ialah 10.

SUDUT KBAT 1 (a) Formula yang boleh digunakan:

n(n + 1)2

= n2 + n2

Bentuk yang kelima:

52 + 52

= 15

(b) Bentuk yang ke-51:

512 + 512

= 1 326

2 x(4x) = 100 4x2 = 100 x = 5 Jumlah umur = 5 + 4(5) = 25

3 √(x – 1)2 + (x – 2)2 = x

√(x2 – 2x + 1 + x2 – 4x + 4) = x

√2x2 – 6x + 5 2 = x2

2x2 – 6x + 5 – x2 = 0 x2 – 6x + 5 = 0 (x – 1)(x – 5) = 0 x = 1 @ 5 Jawapan: 5 sahaja kerana x = 1 akan memberi panjang sisi yang mustahil.

PRAKTIS BAB 2Soalan Objektif 1 C 2 A 3 C 4 C 5 D 6 A 7 A 8 A 9 D 10 C11 A 12 B 13 B 14 C 15 D16 C 17 C

Anjakan Prima Math F4 Jaw 4th.indd 2 9/10/2017 3:50:28 PM

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Soalan Subjektif

1 p2 – 8

4 = p + 1

p2 – 8 = 4p + 4 p2 – 4p – 12 = 0 (p – 6)(p + 2) = 0 p = –2 @ 6

2 6m – 4(m – 2) = 6m2

6m – 4m + 8 = 6m2

6m2 – 2m – 8 = 0 2(3m2 – m – 4) = 0 2(3m – 4)(m + 1) = 0 m = –1 @ 4

3

3 (a) 5x2 + 10x

= 5x(x + 2) (b) x2 – 2xy – (x – y)2

= x2 – 2xy – (x2 – 2xy + y2) = x2 – 2xy – x2 + 2xy – y2

= – y2

4 (a) x2 – 3x + 2 = 0 (x – 1)(x – 2) = 0 x = 1 @ 2 (b) 2y2 – 7y – 4 = 0 (2y + 1)(y – 4) = 0

y = – 12

@ 4

5 k2 = 5 – 8k4

4k2 = 5 – 8k 4k2 + 8k – 5 = 0 (2k + 5)(2k – 1) = 0

k = – 52

@ 12

6 (2t – 2)(t – 2) – 12 = 0 2t2 – 4t – 2t + 4 – 12 = 0 2t2 – 6t – 8 = 0 2(t2 – 3t – 4) = 0 2(t – 4)(t + 1) = 0 t = –1 @ 4 7 (5y – 3)2 = 16y2

(5y – 3)(5y – 3) = 16y2

25y2 – 15y – 15y + 9 = 16y2

9y2 – 30y + 9 = 0 3(3y2 – 10y + 3) = 0 3(3y – 1)(y – 3) = 0

y = 13

@ 3

8 (m – 4)(m + 4)6

= m

m2 + 4m – 4m – 16 = 6m m2 – 6m – 16 = 0 (m – 8)(m + 2) = 0 m = –2 @ 8 9 Luas = (x + 2)(x – 3) 6 = x2 – 3x + 2x – 6 6 = x2 – x – 6 0 = x2 – x – 12 0 = (x – 4)(x + 3) x = –3 @ 4; x = 4

10 Luas = 12

(2x – 1)(x + 3)

152

= 12

(2x2 + 6x – x – 3)

152

= 12

(2x2 + 5x – 3)

2(15) = 2(2x2 + 5x – 3) 30 = 4x2 + 10x – 6 0 = 4x2 + 10x – 36 0 = 2(2x2 + 5x – 18) 0 = 2(2x + 9)(x – 2)

x = – 92

@ 2; x = 2

Tapak = 2x – 1 = 2(2) – 1 = 4 – 1 = 3 cm

BAB 3 Set

3.1 Set 1 (a) Set nombor genap dari 10 hingga 20.

{10, 12, 14, 16, 18, 20} (b) Set lima nombor kuasa dua sempurna

pertama {1, 4, 9, 16, 25} 2 (a) Q = {24, 27, 30, 33, 36, 39}; n(Q) = 6 (b) R = {23, 29, 31}; n(R) = 3 (c) S = {33, 34, 35, 36, 37, 38, 39}; n(S) = 7 (d) T = {4, 8, 12, 16, 20, 24, 28}; n(T) = 7 (e) U = {2, 3, 5}; n(U) = 3 3 (a) (b) (c) (d) (e) 4 (a)

J1

3

9

(b)

Q

54 60

66 72 78

(c)

R

2123 25

27 29

5 (a) 5 (b) 7 (c) 3 6 (a) K f (b) L f (c) M = f 7 (a) Ya (b) Tidak (c) Ya

Latihan Bestari 3.1 1 (a) Set lima nombor perdana yang pertama. (b) Set lima nombor genap yang pertama. 2 (a) K = {21, 23, 25, 27, 29} (b) L = {1, 4, 9, 16, 25, 36, 49} 3 (a) Benar (b) Palsu 4 (a)

13T

15 17

19 21 23

25 27 29

31 33

(b)

U

A I

5 (a) Y = {a, e, h, k, s, t, y} (b) 7 6 (a) F = f (b) G f 7 (a) Serupa (b) Tak serupa 8 (a) h = 4 (b) u = 8

3.2 Subset, Set Semesta dan Set Pelengkap 1 (a) P Q (b) P Q (c) P Q (d) P Q (e) P Q 2 (a)

ML

j

k

lm

a bc de f

(b)

P Q

15

9 6

8

103

2

4 7

(c)

P0

2

4

1 3

5

(d)

Q

(e)

R

1

4 6

5

2 3

7

3 (a) { }, {3}, {4}, {5}, {3, 4}, {3, 5}, {4, 5}, {3, 4, 5}

(b) { }, {w}, {x}, {w, x} 4 (a) Q9 = {5, 7, 8} (b) Q9 = {2, 4, 6, 8, 10, 12, 14, 16, 18} (c) Q9 = {20, 30, 50}

Latihan Bestari 3.2 1 (a) (b) (c) (d)

Anjakan Prima Math F4 Jaw 4th.indd 3 9/10/2017 3:50:29 PM

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2 (a)

P

Qw

z

xy

(b)

R

S

3 9 15

21

6

18

12

24

(c)

T

U31

9 2

11

13

57

3 (a) bilangan subset = 4 { }, {3}, {4}, {3, 4} (b) bilangan subset = 8 { }, {a}, {b}, {c}, {a, b},

{a, c}, {b, c}, {a, b, c} (c) bilangan subset = 2 { }, {s}

4 (a)

6

10

14

18

22 26

4 8

12 16 20

24 28

Q

(b)

T1 7

9

2

8

4 63

5

5 (a) R9 = {10, 20, 30, 40} (b) R9 = {10, 15, 20, 25, 30, 35, 40, 45}

3.3 Operasi ke atas Set 1 (a) {4, 5} (b) {4, 5} 2 (a)

Q

P

(b)

R

Q

(c)

R

P Q

3 (a) {1, 2, 3, 5, 6, 7, 9, 10} (b) {7, 13, 17} 4 (a) {11, 13, 16, 17, 19} (b) {11, 12, 13, 14, 16, 17, 18, 19} 5 (a)

ST

(b)

US

T

(c)

S

TU

6 (a) {1, 3, 5, 9, 11} (b) {15} 7 (a) {66, 88} (b) {a, c, e, g, i, k} 8 (a)

R

P Q

(b)

�Q

R

P

(c)

R

QP

9 (a) W (X Y) (b) W9 Y X (c) W (Y9 X)

Latihan Bestari 3.3 1 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} P = {1, 3, 5, 7, 9} Q = {1, 2, 4} R = {1, 4, 9} (a) P R = {1, 9} (b) (Q R)9 = {2, 3, 5, 6, 7, 8, 9, 10} P (Q R)9 = {3, 5, 7, 9} 2 (a) P Q = f (b) Q R = Q (c) P Q R = f 3 = {T, R, A, N, S, F, O, M, I} S = {A, O, I} T = {F, O, R, M, A, T} U = {N, A, T, I, O} (a) S U = {N, A, T, I, O} (b) (S T U) = {F, O, R, M, A, T, I, N} (S T U)9 = {S} 4 (a) P Q R

P Q

R

(b) P R Q9

P

Q

R

5

BF

575 40

= 5

SUDUT KBAT 1 A 2 (a)

ξBola sepak

10 6

12

7

Catur

(b) 12

PRAKTIS BAB 3Soalan Objektif 1 D 2 A 3 C 4 A 5 B 6 A 7 D 8 A 9 A 10 D11 C 12 D 13 A 14 C

Soalan Subjektif 1 (a) Q f Q = {2, 3, 5, 7, 11} n(Q) = 5 (b) Q = f (c) Q = f 2 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} P = {3, 6, 9, 12} Q = {1, 3} R = {1, 2, 5, 10} (a) (i) P9 = {1, 2, 4, 5, 7, 8, 10, 11} (ii) Q9 = {2, 4, 5, 6, 7, 8, 9, 10, 11, 12} (b) (i) n(R9) = 8 (ii) n(P Q)9 = 11 3 (a) 3 (b) 8 4 (a) P Q R (b) P (Q9 R) 5 (a)

YS

X

(b)

S

Y

X

6 (a) f (b) K (c) H 7 (a)

RQ

P

(b)

R

P Q

8

PR

Q

9

P QR

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BAB 4 Penaakulan Matematik

4.1 Pernyataan 1 (a) Ya, palsu (b) Bukan, palsu (c) Ya, benar 2 (a) , (b) . (c) . (d) . 3 (a) Pernyataan benar: 3 + 6 . 8 Pernyataan palsu: 3 + 8 , 6 (b) Pernyataan benar: 8 4 2 = 4 Pernyataan palsu: 4 4 2 = 8 (c) Pernyataan benar: –1 3 –5 , 10 Pernyataan palsu: –1 3 10 . –5

Latihan Bestari 4.1 1 (a) Bukan pernyataan (b) Pernyataan (c) Pernyataan (d) Pernyataan (e) Bukan pernyataan (f) Pernyataan 2 (a) , (b) . 3 (a) Pernyataan benar: 2 + (–2) , 5 Pernyataan palsu: –10 – 2 . 5 (b) Pernyataan benar: 90 – 34 , 67 Pernyataan palsu: 90 – 34 . 67 (c) Pernyataan benar: 3 + (–3) = 0 Pernyataan palsu: 7 – 3 . 5

4.2 Pengkuantiti “Semua” dan “Sebilangan” 1 (a) Semua (b) Sebilangan (c) Sebilangan 2 (a) Palsu (b) Palsu (c) Benar 3 (a) Ya. Semua formula bagi luas sebarang

segi empat tepat ialah “panjang × lebar” adalah benar.

(b) Tidak. “Semua sekolah mempunyai cikgu lelaki dan perempuan” adalah tidak benar. 4 (a) Sebilangan nombor perdana adalah

gandaan 3. (b) Sebilangan kereta Proton Saga berwarna hijau. (c) Semua heksagon mempunyai 6 sisi.

Latihan Bestari 4.2 1 (a) Semua pentagon mempunyai 5 sisi. (b) Sebilangan arnab berwarna putih. 2 (a) Benar (b) Palsu (c) Palsu 3 (a) Ya (b) Tidak 4 (a) Sebilangan televisyen dibuat dalam Malaysia. (b) Semua rombus adalah sisi empat.

4.3 Operasi ke atas Pernyataan 1 (a) 9 bukan nombor perdana. (benar) (b) 5 darab m tidak sama dengan m5. (benar) (c) Bukan semua nombor genap boleh dibahagi dengan 2. (palsu)

2 (a) (i) {9} ialah subset bagi nombor ganjil. (ii) {9, 21} ialah subset bagi nombor

ganjil. (b) (i) 3 ialah nombor perdana. (ii) 3 ialah nombor ganjil. (c) (i) 1

5 ialah pecahan.

(ii) 1

5 = 0.2

3 (a) 81 boleh dibahagi dengan 9 dan 3. (b) 0 . (–3 + 1) dan –9. (c) Kuda dan lembu ada 4 kaki. 4 (a) (i) Semua integer positif lebih

daripada 1. (ii) Semua integer negatif kurang

daripada 1. (b) (i) Pentagon ada 5 sisi. (ii) Heptagon ada 7 sisi. (c) (i) –10 + 5 , 1 (ii) 1 , 10 – 5 5 (a) 10 atau 8 adalah lebih daripada –4. (b) 4 ialah gandaan 2 atau 4. (c) Dekagon atau segi tiga ialah poligon. 6 (a) Benar (b) Benar (c) Palsu 7 (a) Palsu (b) Benar (c) Palsu

Latihan Bestari 4.3 1 (a) Heksagon tidak mempunyai 7 sisi. (benar) (b) Bukan semua pecahan mempunyai nilai

kurang daripada 1. (benar) (c) Singa bukan haiwan liar. (palsu) (d) Kuching bukan ibu negeri Sarawak.

(palsu) 2 (a) Kaki manusia digunakan untuk berjalan. Kaki manusia digunakan untuk berlari. (b) x = 1 ialah punca bagi persamaan x2 – 3x + 2 = 0. x = 2 ialah punca bagi persamaan x2 – 3x + 2 = 0. 3 (a) Dia hendak menyanyi atau menulis. (benar) (b) –1 . –2 dan (–2 + (–8)). (benar) (c) –2 lebih daripada –10 atau –20. (benar) 4 (a) Benar (b) Palsu (c) (i) Benar (ii) Benar

4.4 Implikasi 1 (a) Antejadian : x = 1 atau x = –2 Akibat : x2 + x – 2 = 0 (b) Antejadian : 3 , 7 Akibat : 32 , 72

(c) Antejadian : y = 9

Akibat : √y = 3 2 (a) Implikasi 1 : Jika 32 , 42, maka 3 , 4 Implikasi 2 : Jika 3 , 4, maka 32 , 42

(b) Implikasi 1 : Jika m – n . 0, maka m . n Implikasi 2 : Jika m . n, maka m – n . 0 (c) Implikasi 1 : Jika n , 7, maka 4n , 28 Implikasi 2 : Jika 4n , 28, maka n , 7 3 (a) Jika x ialah gandaan 2, maka x ialah nombor genap

(b) Jika r = 5, maka 5r = r2. 4 (a) –2 + x = 0 jika dan hanya jika x = 2. (b) r2 = 36 jika dan hanya jika r = 6. 5 (a) Jika p , 3, maka p , 0. (palsu) (b) Jika 3x – 5 = –11, maka x = –2. (benar)

Latihan Bestari 4.4 1 (a) Antejadian : Panjang sebuah segi empat

sama ialah 2 cm. Akibat : Perimeternya ialah 8 cm. (b) Antejadian : 4 . 2 Akibat : 42 – 22 . 0 (c) Antejadian : x = 0° Akibat : kos x = 1 2 (a) Implikasi 1 : Jika U . V, maka –U , –V Implikasi 2 : Jika –U , –V, maka U . V (b) Implikasi 1 : Jika dia boleh menyambung pengajian-

nya, maka dia mempunyai wang. Implikasi 2 : Jika dia mempunyai wang,

maka dia boleh menyambung pengajian-

nya. 3 (a) Jika x , y, maka x – y , 0

(b) Jika ab . 1, maka a . b

4 (a) x(x – 1) = 0 jika dan hanya jika x = 0 atau x = 1.

(b) 2 3 3 = 6 jika dan hanya jika 62 = 3.

5 (a) Jika x , 10, maka x . 12. (palsu) (b) Jika y2 = 81, maka y = 9. (palsu) (c) Jika set P = set Q, maka

P Q = P Q. (benar) (d) Jika X Y, maka X Y = Y. (benar)

4.5 Hujah

1 (a) Premis 1 : Semua burung mempunyai 2 sayap.

Premis 2 : Burung pipit mempunyai 2 sayap.

Kesimpulan : Burung pipit ialah sejenis burung.

(b) Premis 1 : Semua sekolah mempunyai pelajar dan guru.

Premis 2 : Sekolah Menengah Sri Permai ialah sebuah sekolah.

Kesimpulan : Sekolah Menengah Sri Permai mempunyai pelajar dan guru.

(c) Premis 1 : Semua segi tiga mempunyai 3 bucu. Premis 2 : WXY ialah sebuah segi tiga. Kesimpulan : WXY mempunyai 3 bucu. 2 (a) Motosikal Honda ada dua roda. (b) x 5. 3 (a) Jika n ialah nombor genap, maka ia boleh

dibahagi dengan 2. (b) Pokok kelapa ialah sejenis pokok.

Latihan Bestari 4.5 1 Premis 1 : Semua ikan boleh berenang. Premis 2 : Ikan yu ialah sejenis ikan. Kesimpulan : Ikan yu boleh berenang.

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2 (a) Jika x = 2, maka x2 = 4. x 2. Maka, x2 4. (b) Jika 18 boleh dibahagi dengan p, maka p

ialah faktor bagi 18. p bukan faktor bagi 18. Maka, 18 tidak boleh dibahagi dengan p.

3 (a) Jika jejari sebuah bulatan ialah r, maka luas bulatan itu ialah πr2.

(b) P Q Q.

4.6 Deduksi dan Aruhan 1 (a) Aruhan (b) Deduksi (c) Aruhan 2 (a) Maka, hasil tambah sudut pedalaman

ABCDE ialah 540°. (b) Maka, unta melahirkan anak.

3 (a) 12 [(n + 1)], di mana n = 1, 2, 3, 4, …

(b) 2n2 + 5, di mana n = 1, 2, 3, 4, … (c) 2n2 – 1, di mana n = 1, 2, 3, 4, … (d) 5 – 2n, di mana n = 1, 2, 3, 4, …

Latihan Bestari 4.6 1 (a) Deduksi (b) Aruhan 2 2 + n3, di mana n = 1, 2, 3, 4, …

SUDUT KBAT 1 (a) 2n + 1, n = 1, 2, 3, 4, … (b) 2(40) + 1 = 81

PRAKTIS BAB 4Soalan Subjektif 1 (a) 3 + 5 . 7 (b) {2, 3} {2, 3, 4} 2 (a) Benar (b) Palsu 3 (a) Benar (b) Palsu 4 (a) Semua integer positif adalah lebih besar

daripada integer negatif. (benar) (b) Semua harimau bintang boleh memanjat.

(benar) 5 (a) Palsu (b) Benar (c) Palsu (d) Benar 6 (a) Maka x = 150° bukan sudut tirus. (b) Maka x = 20° ialah sudut tirus. 7 7 + 4n, di mana n = 1, 2, 3, 4, … 8 (a) Setiap sudut pedalaman segi tiga ABC

bukan 60°. (b) 6 ialah faktor bagi 12.

BAB 5 Garis Lurus

5.1 Kecerunan Garis Lurus 1 (a) Jarak mencancang = 4 Jarak mengufuk = 6 (b) Jarak mencancang = 0 Jarak mengufuk = 7

Latihan Bestari 5.1 1 (a) Jarak mencancang = 3 Jarak mengufuk = 4 (b) Jarak mencancang = 3 Jarak mengufuk = 0 (c) Jarak mencancang = 2 Jarak mengufuk = 4

(d) Jarak mencancang = 3 Jarak mengufuk = 1

2 (a) 43

(c) 24

= 12

(b) – 32

(d) 30

=

3 8Jarak mengufuk

= 4

Jarak mengufuk = 8

4 = 2

5.2 Kecerunan Garis Lurus dalam Sistem Koordinat Cartes

1 (a) 6 – 2–4 – 0

= 4–4

= –1

(b) –5 – 20–4 – 1

= –25–5

= 5

(c) 8 – 26 – (–3)

= 69

= 23

(d) –6 – 2–3 – 1

= –8–4

= 2

(e) 2 – 112 – 5

= –9–3

= 3

2 (a) 4 – (–1)2 – (–3)

= 55

= 1

(b) 1 – (–2)–3 – 2

= 3–5

= – 35

(c) 4 – 26 – 2

= 24

= 12

(d) 5 – 00 – 4

= – 54

(e) 6 – 2–2 – (–10)

= 48

= 12

3 (a) –3 – 1

2 – p = 2

–4 = 4 – 2p 2p = 8 p = 4

(b) 8 – p2 – 6

= –3 8 – p = 12 p = –4

(c) p – 0 = – 1 1 – (–2) 2 2p = –3 p = – 3 2

Latihan Bestari 5.2 1 (a) P(–2, 4), Q(4, –2)

Kecerunan PQ = –2 – 44 – (–2)

= –66

= –1

(b) R(–6, –1), S(2, 3)

Kecerunan RS = 3 – (–1)2 – (–6)

= 48

= 12

2 (a) (1, 3) dan (2, –8)

Kecerunan = –8 – 32 – 1 = – 11

1 = –11

(b) (2, 5) dan (1, 7)

Kecerunan = 7 – 51 – 2 = 2

–1 = –2

(c) (3, 2) dan (2, 4)

Kecerunan = 4 – 22 – 3 = 2

–1 = –2

(d) (–2, 1) dan (0, 6)

Kecerunan = 6 – 10 – (–2)

= 52

= 52

3 (a) mOP = 3 – 03 – 0

= 1

(b) mPQ = 3 – 03 – 6 = 3

–3 = –1

5.3 Pintasan

1 (a) Pintasan-x = –5, Pintasan-y = 3 (b) Pintasan-x = –5, Pintasan-y = –4

2 (a) – –3–3

= –1 (b) – 2–4

= 12

3 (a) – 6x

= –3 (b) – 3x = 1

2 –6 = –3x 6 = –x x = 2 x = –6

4 (a) – y–2 = 3

4 (b) – y3

= –4

–4y = –6 –y = –12

y = 32 y = 12

Latihan Bestari 5.3 1 No. Titik R Titik S Pintasan-x Pintasan-y Kecerunan

(a) (0, 3) (1, 0) 1 3 –3

(b) (3, 0) (0, –6) 3 –6 2

(c) (0, 5) (–10, 0) –10 5 12

2 (a) –3 = – Pintasan-y2

Pintasan-y = 6

(b) – 12

= – Pintasan-y4

Pintasan-y = 2

(c) 25

= – 10Pintasan-x

Pintasan-x = – 10 3 52

= –25

5.4 Persamaan Garis Lurus

1 (a)

x 0

32

y –3 0

y

x0

–3

32

(b)

x 0 –3

y 9 0

y

x0

9

–3

2 (a) Tidak (b) Ya (c) Ya

3 (a) y = – 14 x + 3

4

(b) y = 2

3 x – 6

4 (a) 2x + 3y = 1 (b) 4y = 8x – 4 y = 2x – 1 m = 2, c = –1

3y = –2x + 1

y = – 23 x + 1

3

m = – 23 ; c = 1

3

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5 (a) 1 = 3(–4) + c (b) 4 = (8) + c

c = 4 – 6 c = –2 y = x – 2

3 4

3 4

c = 1 + 12 c = 13 y = 3x + 13

6 (a) m = 4 – 12 – 3 (b) m =

= 3

–6 = – 12

3 = – 12

(–1) + c

c = 3 – 12

= 52

y = – 12

x + 52

3 – 0–1 – 5

= 3–1

= –3 4 = –3(2) + c c = 4 + 6 = 10 y = –3x + 10

7 (a) y = – 12 x + 2 → ①

y = x – 1 → ② Gantikan ② ke dalam ①

x – 1 = – 12 x + 2

2x – 2 = –x + 4 3x = 6 x = 2 Gantikan x = 2 ke dalam ② y = 2 – 1 = 1 Titik persilangan ialah (2, 1) (b) y = x + 1 → ① 2y = x + 4 → ② Gantikan ① ke dalam ② 2(x + 1) = x + 4 2x + 2 = x + 4 x = 2 Gantikan x = 2 ke dalam ① y = 2 + 1 = 3 Titik persilangan ialah (2, 3) (c) y = –3x + 2 → ① 2y = 4x + 8 → ② Gantikan ① ke dalam ② 2(–3x + 2) = 4x + 8 –6x + 4 = 4x + 8 10x = –4

x = – 25

Gantikan x = – 25 ke dalam ①

y = –3– 25 + 2

y = 6 5 + 2

y = 3 15

Titik persilangan ialah – 25 , 3

15

(d) 2y = 3x – 5 → ① 4y = 3x + 5 → ② ① – ② –2y = –10 y = 5 Gantikan y = 5 ke dalam ① 2(5) = 3x – 5 3x = 10 + 5 x = 5 Titik persilangan ialah (5, 5).

Latihan Bestari 5.4 1 (a) y = 3x

x 0 1

y 0 3

y

x

21

3

0 1

(b) y = – x2

+ 2

x 0 2

y 2 1

y

x

21

0 21 3

(c) y = –3x + 2

x 0 1

y 2 –1

y

x

21

–10 21 3

2 (a) y = –2x + 3; (0, 3) Gantikan x = 0 dan y = 3 ke dalam y = –2x + 3 KIRI = y = 3 KANAN = –2(0) + 3 = 3 Oleh kerana KIRI = KANAN, titik (0, 3)

terletak di atas garis lurus y = –2x + 3 (b) y = –2x + 3; (4, –3) Gantikan x = 4 dan y = –3 ke dalam y = –2x + 3 KIRI = y = –3 KANAN = –2(4) + 3 = –5 Oleh kerana KIRI KANAN, titik (4, –3)

tidak terletak di atas garis lurus y = –2x + 3 (c) y = –2x + 3; (2, –4) Gantikan x = 2 dan y = –4 ke dalam y = –2x + 3 KIRI = y = –4 KANAN = –2(2) + 3 = –1 Oleh kerana KIRI KANAN, titik (2, –4)

tidak terletak di atas garis lurus y = –2x + 3 3 (a) m = –1, c = 3 y = –1(x) + 3 y = –x + 3

(b) m = 12 , c = 2

y = 12 (x) + 2

y = 12 x + 2

4 (a) 2y = 4x + 3

y = 2x + 32

m = 2, c = 32

(b) 2x – y = 1 y = 2x – 1 m = 2, c = –1

(c) y = –2x + 9 m = –2, c = 9

5 (a) Selari dengan paksi-x bermaksud garis lurus itu terletak pada koordinat-y titik y = 1

(b) Selari dengan paksi-y bermaksud garis lurus itu terletak pada koordinat-x titik x = 3

6 (a) m = 3, (1, 2) (b) m = –1, (3, 0) 0 = –1(3) + c c = 3 y = –x + 3

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

7 (4, –1) dan (1, 3)

m = 3 – (–1)1 – 4

= 4–3

y = – 43 x + c

3 = – 43 (1) + c

c = 3 + 43

= 13

3

y = – 4

3 x + 133

8 (a) y = x

2 … ①

y + x = 3 … ②

y = –x + 3 … ③ Gantikan ③ ke dalam ①

–x + 3 = x2

–2x + 6 = x

–3x = –6 x = 2 Gantikan x = 2 ke dalam ② y + 2 = 3 y = 1 Titik persilangan ialah (2, 1). (b) y = x + 2 … ① y – 4x + 4 = 0 … ②

Gantikan ① ke dalam ②

x + 2 – 4x + 4 = 0 –3x + 6 = 0 3x = 6 x = 2 Gantikan x = 2 ke dalam ② y – 4(2) + 4 = 0 y = 8 – 4 = 4 Titik persilangan ialah (2, 4).

5.5 Garis Selari 1 (a) Tidak (b) Ya (c) Tidak

2 (a) m = –1 3y = –px – 8

y = – p3 x – 8

3 –1 = –

p3

p = 3

(b) m = –

2y = – px + 7

y = – x + 72

– = –

4p = 6 p =

34

34

p2p2

32

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(c) m = –3 p = –3 3 (a) m = 3 –2 = 3(–2) + c c = –2 + 6 c = 4 ∴ y = 3x + 4 (c) m = – 3

2 6 = – 3

2(–4) + c

c = 0 ∴ y = – 3

2x

4 (a) m = 12

2 = 12 (0) + c

c = 2 y = 1

2 x + 2

(b) m = –1 4 = –1(0) + c c = 4 ∴ y = –x + 4

Latihan Bestari 5.5 1 (a) 2y = x – 10

y = 12 x – 5

m = 12

5y = tx + 6

y = tx5 – 6

5 t

5 = 12

(b) y = –2x + 1 m = –2 2y = tx + 1

y = x +

= –2

t = –4

12

t2t

2

2t = 5 t = 5

2

2 (a) M(3, –2); 4y = 12x – 9

y = 3x – 94

m = 3

–2 = 3(3) + c c = –2 – 9 = –11 y = 3x – 11 (b) M(–1, 2); 4x – 6y = 1 6y = 4x – 1 y = 2

3 x – 16

m = 2

3

2 = 23 (–1) + c

6 = –2 + 3c

3c = 8

c = 8

3

y = 23 x + 8

3 3 (a) R(0, 0), S(5, 3) m = 3 – 0

5 – 0 = 35

(b) 3 = 3

5 (5) + c

3 = 3 + c c = 0 y = 3

5 x

(c) P(0, 3), m = 35

3 = 35 (0) + c

c = 3

y = 3

5 x + 3

0 = 35 x + 3

35 x = –3

3x = –15 x = –5 Pintasan-x = –5

SUDUT KBAT 1 (a) y = 70x y = 30x + 80 (b) Hari ke-2. Bayaran sewa = RM40 (c) Syarikat Abu, RM290

PRAKTIS BAB 5Soalan Objektif 1 D 2 D 3 B 4 D 5 A 6 A 7 C 8 C 9 C

Soalan Subjektif 1 (a) O(0, 0), T(2, 2) Kecerunan OT = 2 – 0

2 – 0

= 2

2 = 1 (b) T(2, 2), R(6, 0) Kecerunan PR = 0 – 2

6 – 2

= –2

4

= – 12

2 (2, 6) dan (–3, 4)

m = 4 – 6–3 – 2

= –2

–5

= 25

6 = 2

5 (2) + c

c = 6 – 45

= 26

5

y = 2

5x + 26

5 3 (1, 2); 2x + 3y + 2 = 0 3y = –2x – 2

y = – 23

x – 23

m = – 2

3

2 = – 2

3 (1) + c

c = 2 + 23

= 83

y = – 2

3 x + 83

4 (a) P(–2, 0), T(0, 6)

m = 6 – 00 – (–2)

= 6

2 = 3

(b) PT = √62 + (–2)2

= √36 + 4

= √40 = 6.32 k – 0 = 6.32 k = 6.32

5 (a) m = 34 , pintasan-x = 6

34 = –

pintasan-y6

pintasan-y = – 6 3 3

4 = – 9

2

(b) m = 3

4 , c = –2

34 = – –2

pintasan-x

pintasan-x = 4 3 2

3 = 8

3 6 (a) R(4, 0); y = 2x – 4 m = 2 0 = 2(4) + c c = –8 y = 2x – 8 (b) R(2, –2); y = –3x + 1 m = –3 –2 = –3(2) + c c = –2 + 6 = 4 y = –3x + 4

7 (a) 8x + 6y = 48 pintasan-y, x = 0 6y = 48 y = 8 (b) 8x + 6y = 48 pintasan-x, y = 0 8x = 48 x = 6 pintasan-x = 6 (c) 6y = –8x + 48 y = – 8

6x + 8

m = – 86

= – 4

3

8 (a) y = 2x + 1 … ①

4x + y = 7 … ② Gantikan ① ke dalam ② 4x + (2x + 1) = 7 4x + 2x = 6 6x = 6 x = 1 Gantikan x ke dalam ② 4(1) + y = 7 y = 7 – 4 = 3 Titik persilangan ialah (1, 3)

9 (a) m = 35

y = 3

5x + c

0 = 35

(–4) + c

c = 125

y = 3

5 x + 125

(b) m = –3 –4 = –3(0) + c c = –4 ∴ y = –3x – 4

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(b) R(x, y), O(0, 0), m = 35

y – 0x – 0

= 35

R(5, 3)

y = 3

5x + c

3 = 35

(5) + c

3 = 3 + c c = 0

y = 35

x

(c) R(5, 3), S(10, 0)

m =

3 – 0

5 – 10

= – 35

P(h, 8), S(10, 0)

m = – 3

5

0 = – 35

(10) + c

c = 6

y = – 35

x + 6

(d) P(h, 8), S(10, 0), m = – 35

m =

0 – 8

10 – h = – 3

5 –40 = –30 + 3h 3h = –10

h = – 103

BAB 6 Statistik

6.1 Selang Kelas 1 (a)

Panjang (cm)85 – 8990 – 9495 – 99

100 – 104

(b)

Jisim (kg)30 – 3435 – 3940 – 44

2

Had Had Sempadan Sempadan Saiz bawah atas bawah atas kelas

45 49 44.5 49.5 5

50 54 49.5 54.5 5

51 60 50.5 60.5 10

1.1 1.5 1.05 1.55 0.5

3.0 3.9 2.95 3.95 1.0

5.0 5.4 4.95 5.45 0.5

3 (a) Saiz selang kelas = 10

Selang Kekerapan kelas

11 – 20 2 21 – 30 7 31 – 40 6 41 – 50 4 51 – 60 3 61 – 70 8 71 – 80 4 81 – 90 4 91 – 100 2

(b) Saiz selang kelas = 11

Selang Kekerapan kelas

11 – 21 3 22 – 32 11 33 – 43 9 44 – 54 2 55 – 65 1 66 – 76 4

Latihan Bestari 6.1 1

Jejari Had Had Sempadan Sempadan Saiz (cm) bawah atas bawah atas kelas

5 – 9 5 9 4.5 9.5 5

10 – 14 10 14 9.5 14.5 5

15 – 19 15 19 14.5 19.5 5

2

Selang Kekerapan kelas

4.2 – 4.6 1 4.7 – 5.1 0 5.2 – 5.6 1 5.7 – 6.1 4 6.2 – 6.6 7 6.7 – 7.1 12 7.2 – 7.6 6 7.7 – 8.1 8 8.2 – 8.6 4 8.7 – 9.1 7

3 (a) 490 (b) 489 (c) 479.5 (d) 10

6.2 Mod dan Min bagi Data Terkumpul 1 (a) (21 – 25) cm (b) (30 – 34) g 2 (a)

Titik tengah525762

(b)

Titik tengah30.540.550.5

3 (a)

Umur Kekerapan Titik tengah Titik tengah 3 (tahun) Kekerapan

14 – 16 16 15 240

17 – 19 11 18 198

20 – 22 15 21 315

23 – 25 22 24 528

26 – 28 10 27 270

29 – 31 16 30 480

Jumlah 90 Jumlah 2 031

Min = 2 03190

= 22.6

(b)

Masa Kekerapan Titik Titik tengah 3 (minit) tengah Kekerapan

10 – 14 6 12 72

15 – 19 7 17 119

20 – 24 7 22 154

25 – 29 16 27 432

30 – 34 3 32 96

35 – 39 1 37 37

Jumlah 40 Jumlah 910

Min = 91040

= 22.75

Latihan Bestari 6.2 1 (a) 21 – 30 (b) 150 – 159 2 (a)

Titik tengah23283338

2 (b)

Titik tengah12172227

3 (a) Min = (17 3 5) + (22 3 4) + (27 3 22)31

= 24.74

(b) Min = (3 3 6) + (8 3 3) + (13 3 7)16

= 8.31

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6.3 Histogram 1 (a)

Sempadan Sempadan bawah atas 40.5 45.5 45.5 50.5 50.5 55.5 55.5 60.5 60.5 65.5

1

Markah

Kek

erap

an

2

3

4

5

6

040.5 45.5 50.5 55.5 60.5 65.5

(b)

Sempadan Sempadan bawah atas 1.5 6.5 6.5 11.5 11.5 16.5 16.5 21.5 21.5 26.5

1

Masa (minit)

Kek

erap

an

2

3

4

5

6

01.5 6.5 11.5 16.5 21.5 26.5

2 (a) (i) 3

(ii) Min = (6 3 4) + (9 3 2) + (12 3 8)

+ (15 3 10) + (18 3 6) 30

= 13.2

(b) (i) (155 – 159) cm (ii) 40

50 3 100% = 80%

Latihan Bestari 6.3 1

Panjang (cm)

Kek

erap

an

1

2

3

4

5

6

7

8

9

029.5 39.5 49.5 59.5 69.5 79.5 89.5

6.4 Poligon Kekerapan 1 (a)

2

Umur (tahun)

Kek

erap

an

4

6

8

10

011.5 15.5 19.5 23.5 27.5 31.5 35.5

(b)

1

Tinggi (cm)

Kek

erap

an

2

3

4

5

0120.5 131.5 142.5 153.5 164.5 175.5

(c)

5

Berat (kg)

Kek

erap

an

10

15

20

25

015.5 20.5 25.5 30.5 35.5 40.5 45.5

Jisim (kg)

(d)

20

Skor

Kek

erap

an

40

60

80

100

026.5 30.5 34.5 38.5 42.5 46.5 50.5

2 (a)

1

Skor

Kek

erap

an

2

3

4

5

6

040.5 50.5 60.5 70.5 80.5 90.5100.5

(b)

1

Markah

Kek

erap

an

2

3

4

5

6

05.5 10.5 15.5 20.5 25.5 30.5 35.5 40.5

Latihan Bestari 6.4 1 (a) (46 – 50) kg (b) 6

6.5 Kekerapan Longgokan 1 (a)

Jisim (g) 41 – 50 51 – 60 61 – 70 71 – 80 Kekerapan 3 4 3 8

Sempadan 50.5 60.5 70.5 80.5 atas

Kekerapan 3 7 10 18 longgokan

(b)

Jarak (km) 11 – 15 16 – 20 21 – 25 26 – 30 Kekerapan 1 6 6 2

Sempadan 15.5 20.5 25.5 30.5 atas

Kekerapan 1 7 13 15 longgokan

2 Masa 11 – 20 21 – 30 31 –40 41 – 50 51 – 60 (minit)

Kekerapan 5 7 8 7 3

Sempadan 20.5 30.5 40.5 50.5 60.5 atas

Kekerapan 5 12 20 27 30 longgokan

Masa (minit)

Kek

erap

an lo

nggo

kan

5

10

15

20

25

30

010.5 20.5 30.5 40.5 50.5 60.5

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3 Skor 10 – 19 20 – 29 30 – 39 40 – 49 50 – 59

Kekerapan 3 9 6 4 3

Sempadan 19.5 29.5 39.5 49.5 59.5 atas

Kekerapan 3 12 18 22 25 longgokan

Skor

Kek

erap

an lo

nggo

kan

5

10

15

20

25

09.5 19.5 29.5 39.5 49.5 59.5

Latihan Bestari 6.5 1

Skor Kekerapan Sempadan atas

Kekerapan longgokan

50 – 59 4 59.5 460 – 69 6 69.5 1070 –79 17 79.5 2780 – 89 8 89.5 3590 – 99 5 99.5 40

2 Harga (RM) Kekerapan Sempadan

atasKekerapan longgokan

41 – 50 8 50.5 851 – 60 13 60.5 2161 –70 25 70.5 4671 – 80 13 80.5 5981 – 90 7 90.5 66

91 – 100 4 100.5 70

Harga (RM)

Kek

erap

an lo

nggo

kan

10

20

30

40

50

60

70

040.5 50.5 60.5 70.5 80.5 90.5 100.5

6.6 Sukatan Serakan 1 (a) 30 (b) 30 (c) 19.1 2 (a) Median = 46 jam Kuartil pertama = 43 jam Kuartil ketiga = 55 jam Julat antara kuartil = 12 jam (b) Median = 55 Kuartil pertama = 46 Kuartil ketiga = 62 Julat antara kuartil = 16 (c) Median = 17.5 tahun Kuartil pertama = 13 tahun Kuartil ketiga = 23.5 tahun Julat antara kuartil = 10.5 tahun

3 (a) (i) 60 (ii) 14 (b) (i) 25 (ii) 17.5 (iii) 5 (c) (i) 60 kg (ii) 67 kg

Latihan Bestari 6.6 1 (a) Julat = 19 – 3 = 16

(b) Julat = 25 + 292 – 10 + 14

2 = 27 – 12 = 15

2 (a) Median = 30.5 (b) Kuartil pertama = 25.5 Kuartil ketiga = 36.5 Julat antara kuartil = 36.5 – 25.5 = 11

SUDUT KBAT

1 (b) (i) min =

(7 × 12) + (12 × 20) + (17 × 20) + (22 × 33) + (27 + 15)

100 = 17.95

(ii) Peratusan = 15100 3 100 = 15%

(b) 15 × RM10 = RM150

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

Soalan Subjektif 1 (a)

Selang Kekerapan kelas

140 – 149 5 150 – 159 7 160 – 169 5 170 – 179 2 180 – 189 1

(b) (i) 10 (ii) 150 – 159 2 (a)

Skor

Kek

erap

an

1

2

3

4

5

6

0 0.5 5.5 10.5 15.5 20.5 25.5 30.5

(b) (i) 21 – 25 (ii) Skor min

=

(3 3 2) + (8 3 4) + (13 3 3) + (18 3 5) + (23 3 6) + (28 3 5)

25

= 44525

= 17.8

3 (a) 32 (b) (61 – 65) kg (c) Jisim min

=

(48 3 5) + (53 3 8) + (58 3 6) + (63 3 10) + (68 3 3)

32 = 1 846

32 = 57.6 kg

4 (a) 32 (b) Nilai min

=

(455.5 3 3) + (465.5 3 7) + (475.5 3 8) + (485.5 3 5) + (495.5 3 5) + (505.5 3 4)

32

= 15 35632

= RM479.88

5 (a)

Masa (jam)

Kek

erap

an

1

2

3

4

5

6

7

010.5 20.5 30.5 40.5 50.5 60.5 70.5 80.5 90.5

(b) Min

=

(25.5 3 5) + (35.5 3 3) + (45.5 3 7) + (55.5 3 6) + (65.5 3 4) + (75.5 3 5)

30

= 1 52530

= 50.83 jam 6 (a) Bilangan kasut yang dijual = 10 + 13 + 17 + 8 + 2 = 50 (b) 6 7

Kekerapan 10 26 43 50 61 70 longgokan

8 (a) Julat = 19 – 3 = 16 Median = 6 (b) Julat = 4.1 – 0.5 = 3.6 Median = 1.2 9 (a) 60 (b) Median = 60 10 (a)

Kekerapan 7 19 38 52 60 longgokan

(b)

Tempoh (jam)

Kek

erap

an lo

nggo

kan

10

20

30

40

50

60

09.5 14.5 19.5 24.5 29.5 34.5

Kuartil pertama

Median

Kuartil ketiga

(c) (i) Median = 22 (ii) Julat antara kuartil

= 26.5 – 18 = 8.5

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BAB 7 Kebarangkalian I

7.1 Ruang Sampel 1 (a) (i) Mungkin (ii) Tidak mungkin (b) (i) Tidak mungkin (ii) Mungkin (c) (i) Mungkin (ii) Mungkin (d) (i) Mungkin (ii) Tidak mungkin 2 (a) S, P, A, C, E (b) Matematik, Sains, Sejarah 3 (a) S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (5, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} (b) S = {Kepala Kepala, Kepala Ekor, Ekor Kepala, Ekor Ekor}

Latihan Bestari 7.1 1 (a) Tidak mungkin (b) Mungkin 2 (a) Guli hijau, guli kuning, guli merah (b) 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21,

22, 23, 24, 25, 26, 27, 28, 29 3 S = {(4, 1), (4, 2), (4, 3), (8, 1), (8, 2), (8, 3),

(16, 1), (16, 2), (16, 3)}

7.2 Peristiwa 1 (a) (i) {3, 6, 9, 12} (ii) {9} (b) (i) {11, 12, 13, 14, 15, 16, 17, 18, 19} (ii) {12, 14, 16, 18} (c) (i) {M, A, G, I, C} (ii) {A, I} (d) (i) {1, 2, 3, 4, 5, 6} (ii) {2, 3, 5}

Latihan Bestari 7.2 1 (a) W = {2, 4, 6} (b) X = {A, I} (c) Y = {2, 3, 5, 7} 2 {12, 18, 24, 30, 36}

7.3 Kebarangkalian Suatu Peristiwa

1 (a) (i) 411

(ii) 7

11

(b) (i) 35

(ii) 3

5

(c) 71100

(d) (i) 110

(ii) 3

10

2 (a) Bilangan hari hadir lambat

= 25

× 25 = 10 hari

(b) Bilangan pelajar lelaki

= 49

× 45 = 20 orang

(c) (i) Pen merah = 77

– 27

= 57

(ii) Bilangan pen = 10 ÷ 2 × 7 = 35 batang (d) (i) Jumlah guli = 50 ÷ 1 × 4 = 200

Bilangan guli biru = 35

× 200

= 120 biji (ii) Bilangan guli hijau = 200 – 120 – 50 = 30 biji

x = 30200

= 320

Latihan Bestari 7.3 1 Luas segi empat sama = 18 cm 3 18 cm = 324 cm2

Luas bahagian berlorek = 10 cm 3 10 cm = 100 cm2

Kebarangkalian damak di bahagian berlorek

= 100324

= 2581

2

29

SUDUT KBAT 1 (a) 3

8

(b) Jumlah bola = 85

× 30 = 48

m = 48 – 30 = 18 2 Kebarangkalian = 60

300 = 1

5

Bilangan orang lemah = 15

× 2 000

= 400

PRAKTIS BAB 7

Soalan Objektif 1 A 2 B 3 B 4 C 5 D 6 B 7 B 8 C 9 B 10 A

BAB 8 Bulatan III

8.1 Tangen kepada Bulatan 1 (a) Tidak (b) Ya (c) Ya

2 (a)

O

tangen

P

(b)

O

P

tangen

tangen

3 (a) 72° (b) 22° (c) 62°

4 (a) tan θ = 611

= 0.5454 θ = 28.61° (b) POQ = 2 × 72° = 144° θ = 180° – 144° = 36° (c) QOR = 180° – (2 × 28°) = 180° – 56° = 124° θ = 124° ÷ 2 = 62°

Latihan Bestari 8.1 1 CD dan GH 2 POQ = 25° 3 2 = 50° u = 180° – 90° – 50° = 40° 3 OP = √92 + 52 = 10.30 cm 4 QOT = 40° 3 2 = 80° u = 180° – 90° – 80° = 10°

8.2 Sudut di antara Tangen dan Perentas 1 (a) BAQ (b) QTS (c) 50° (d) 94° 2 (a) θ = DEB = 180° – 62° – 71° = 47° (b) θ = QST = (180° – 48°) ÷ 2 = 66° (c) θ = 180° – 62° – 90° = 28° (d) DOE = 2 × 52° = 104° EDO = (180° – 104°) ÷ 2 = 38° θ = BDE = 38° + 28° = 66°

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Latihan Bestari 8.2 1 PQB 2 u = QST = 180° – 52° – 88° = 40° 3 WQO = (180° – 76°) 4 2 = 52° u = 90° – 52° = 38° 4 u = 48° + 43° = 91°

8.3 Tangen Sepunya 1 (a) (i) BE (ii) CAD (b) 7

10 = 9

PC PC = 90

7 cm

(c) kos APB = 8

11

APB = 43.34° u = APB = 43.34° (d) TPQ = 31° + 31° = 62° u = 360° – 90° – 90° – 62° = 118° 2 (a) BC = √122 – 72 = √95 = 9.75 cm (b) (i) PE (ii) DQA @ CPA @ BQA (iii) BC = ED = 21 cm – 10 cm = 11 cm (c) QC = √122 + 42 = √160 = 12.64 cm

PC = 2012

× 12.64 cm = 21.07 cm PQ = PC – QC = 21.07 cm – 12.64 cm = 8.43 cm

Latihan Bestari 8.3 1 PQ = √202 + 22

= 20.10 cm

2 tan 28° = 7AB

AB = 13.17 cm

tan 28° = 12AC

AC = 22.57 cm

BC = 22.57 – 13.17 = 9.40 cm

SUDUT KBAT 1 = 180

360 × 2 × 3.142 × 10 + 180

360 × 2 × 3.142 × 5

+ (2 × √52 + 502 ) = 147.63

PRAKTIS BAB 8Soalan Objektif 1 C 2 A 3 D 4 A 5 B 6 D 7 A 8 B 9 C 10 B11 B 12 C 13 A 14 B 15 C16 A 17 A 18 D 19 C 20 A

BAB 9 Trigonometri II

9.1 Nilai Sinus, Kosinus dan Tangen Suatu Sudut 1 (a) Sukuan III (b) Sukuan IV (c) Sukuan IV (d) Sukuan I (e) Sukuan II 2 (a) (i) (a) 0.8 (b) –0.7 (ii) (a) 0.6 (b) 0.7

(iii) (a) 43

(b) –1 (b) (i) (a) 0.1 (b) –0.9 (ii) (a) –1 (b) –0.4 (iii) (a) –0.1

(b) 94

3 (a) –0.5 (b) –0.75 (c) 0.7 (d) 1.143 (e) –0.8 4 (a) Negatif (b) Positif (c) Negatif

5 (a) 1√2

(b) √3

2

6 (a) 3 tan 45° + 2 sin 90° = 3(1) + 2(1) = 3 + 2 = 5 (b) 4 kos 60° – 2 sin 30° = 4(0.5) – 2(0.5) = 2 – 1 = 1 (c) tan 180° + 6 sin 30° = 0 + 6(0.5) = 0 + 3 = 3 7 (a) 180° – 120° = 60° (b) 200° – 180° = 20° (c) 360° – 333° = 27° 8 (a) sin 284.5° = –sin(360° – 284.5°) = –sin 75.5 = –0.9681 9 (a) 39.92°, 140.08° (b) 78.64°, 281.36° (c) 72.49°, 252.49°

Latihan Bestari 9.1 1 Sukuan IV

2 tan u = 0.60.76

= 0.7895 3 positif

4 tan 240° = tan 60° = √3

30

601

2

3

5 sin u = 1√2

u = 45°

1

145°

45°2

6 = 60°

300°

60°

7 tan 120° 179 = –1.712 8 sin u = 0.1671 u = 9.62° u = 180° + 9.62° = 189.62°

9.2 Graf Sinus, Kosinus dan Tangen 1 (a)

90°0

1

–1

y

180° 270° 360°�

u = 180°

(b)

90°0

1

–1

y

180° 270° 360°�

u = 90°, 270°

(c)

90°0

1

–1

y

180° 270° 360°�

u = 270°

(d)

90°0

1

–1

y

180° 270° 360°�

u = 90°

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(e)

90°0

1

–1

y

180° 270° 360°�

u = 0°, 180°

2 (a) M = 180° (b) y = sin 2x untuk 0° < x < 360°

Latihan Bestari 9.2 1

270°180°90°

y

x0

–1

1

2

90°

y

x0

1

3

270°

y

x0

–1

1

sin u = –1 u = 270°

SUDUT KBAT 1 x = (6 × sin 60°) – (6 × sin 45°) = 0.953 m

PRAKTIS BAB 9Soalan Objektif 1 C 2 B 3 D 4 D 5 A 6 C 7 B 8 D 9 D 10 B11 A 12 C 13 A 14 D 15 C16 A 17 B 18 B 19 A

BAB 10Sudut Dongakan dan Sudut Tunduk

10.1 Sudut Dongakan dan Sudut Tunduk 1 (a) (i) BC

(ii) tan ABC = 512

ABC = 22.61°

(b) (i) PQ

(ii) kos PQR = 1627

PQR = 53.66°

2 (a) 3.47.6 = 0.4474

tan–1 0.4474 = 24.1°

(b) 19.630 = 0.6533

sin–1 0.6533 = 40.8° (c) 11

16 = 0.6875 tan–1 0.6875 = 34.5°

3 (a) 3040 = 0.75

tan–1 0.75 = 36.9°

(b) 3560 = 0.5833

tan–1 0.5833 = 30.26°

(c) x15 = tan 41

x = 0.8693 × 15 = 13.04 m Tinggi tiang AB = 13.04 m + 19 m = 32.04 m

Latihan Bestari 10.1

1 tan u° = 2418

u = 53.13°

2 sin 50° = QV26

QV = 19.92 m

SUDUT KBAT 1 QW = (15 × tan 45°) = 15 m 2 ML = 2(15 × tan 52°) = 38.398 m

PRAKTIS BAB 10Soalan Objektif 1 A 2 D 3 A 4 A 5 C 6 B 7 A 8 B 9 C 10 B11 A 12 B 13 A 14 D 15 B16 A 17 A 18 C

BAB 11Garis dan Satah dalam Tiga Dimensi

11.1 Sudut antara Garis dan Satah 1 (a) (i) ABCD (ii) VAD, VCD (iii) VAB, VBC (b) (i) ABCD, EFGH (ii) ABFE, BCGF, GCDH, AEHD (iii) Tiada

2 (a) (i) BC, CF, BF (ii) AD, DE, BC, CF (b) (i) AB, BF, AF (ii) AF, DF, BF, CF 3 (a) DE, CF (b) AF, DE 4 (a) (i) DAH (ii) DBH (iii) BAG (b) (i) DAE (ii) EBD 5 (a) tan VEW = 9

6 = 1.5 VEW = 56.31°

(b) BD = √122 + 62

= √180 = 13.42 cm

tan EBD = 813.42

= 0.5961 EBD = 30.8°

(c) tan CFB = 76

= 1.667 CFB = 49.4° (d) BE = √202 + 112

= √521 = 22.83 cm

tan EBH = 1422.83

EBH = 31.52°

Latihan Bestari 11.1 1 AB, CD, EF, HG 2 ABV, BCV 3 BF = √42 + 52

= 6.403 cm

tan u° = 76.403

u = 47.55°

4 Anggap Q ialah titik tengah DH AQ = √42 + 72

= 8.062 cm

tan u° = 108.062

u = 51.12°

11.2 Sudut di antara Dua Satah 1 (a) GH (b) CF 2 (a) BEF @ CHG (b) VFE (c) BFC @ AED

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3 (a) tan FBE = 915

= 0.6 FBE = 30.96°

(b) tan FCD = 710

= 0.7 FCD = 35°

(c) tan BDC = 916

= 0.5625 BDC = 29.36°

Latihan Bestari 11.2 1 BCFE, ADFE 2 AD

3 tan u° = 1015

u = 33.7°

4 Anggap N ialah titik tengah HG

tan u° = 106

u = 59.04°

SUDUT KBAT 1 (a) VTU atau UTV

(b) tan VTU = √62 – 42

8 VTU = 29.19˚ atau 29˚ 11′

PRAKTIS BAB 11Soalan Objektif 1 D 2 B 3 C 4 A 5 D 6 C 7 D 8 B 9 D 10 C11 C 12 D 13 B 14 B 15 B16 B 17 A 18 C 19 A 20 D

Soalan Subjektif

1 tan u° = 910

u = 41.99°

2 tan u° = 1712

u = 54.78°

3 DB = √202 + 62

= 20.88 cm

DE = 20.88 3 12

= 10.44 cm

tan u° = 1510.44

u = 55.16°

4 tan u° = 1411

u = 51.84°

5 BD = √142 + 52

= 14.87 cm

tan u° = 714.87

u = 25.21°

6 AC = √152 + 172

= 22.67 cm

tan u° = 822.67

u = 19.44°

7 AC = √202 – 152

= 13.23 cm

tan u° = 13.2317

u = 37.89°

8 tan u° = 920

u = 24.23°

9 CM = √102 + 142

= 17.20 cm

tan u° = 1617.20

u = 42.93°

10 tan u° = 58

u = 32.01°

Penilaian Akhir Tahun

KERTAS 1 1 C 2 A 3 B 4 C 5 C 6 B 7 B 8 B 9 B 10 B11 D 12 D 13 A 14 C 15 D16 A 17 A 18 C 19 C 20 B21 C 22 D 23 C 24 D 25 B26 A 27 A 28 C 29 C 30 A31 A 32 A 33 A 34 C 35 A36 C 37 C 38 B 39 D 40 A

KERTAS 2 1 (4.71 + 2.962) ÷ 0.00035 = 21 920 = 21 900 (kepada tiga angka bererti)

2 1.92 × 10–4

6 500 = 2.95 × 10–8

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

4 2y(y + 3) = 3 + 5y 2y2 + 6y – 3 – 5y = 0 2y2 + y – 3 = 0 (2y + 3)(y – 1) = 0

y = – 32

atau 1

5 (a)

A B

C

ξ

(b)

A B

C

ξ

6 (a) (x + 1) + 8 + x + (2x + 3) + (x + 4) = 46 5x + 16 = 46 5x = 30 x = 6(b) n(Q′) = (x + 1) + 8 + (x + 4) = 2x + 13 = 2(6) + 13 = 25

7 (a) Bilangan guli merah

= 25

× 40 = 16

Bilangan guli hijau = 40 – 8 – 16 = 16

(b) Bilangan guli hijau = 11, jumlah guli = 35. Jadi, kebarangkalian memilih sebiji guli

hijau = 1135

8 (a) x = 180 – (75 + 20) = 85(b) y = 20 + 20 = 40

9 (a) tan 60° = PQ15

PQ = 15 tan 60° = 26 m(b) katakan sudut dongakan titik Q dari titik

S ialah θ.

tan θ = 26 – 1015

= 1.067

θ = 46° 51′

10 (a) ∠VMA

(b) AM = 162 – 82 = 192

tan ∠VMA = 18192

= 1.3

∠VMA = 52° 26′

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11 (a) Persamaan bagi garis lurus RS ialah x = –4.

(b) Kecerunan bagi garis lurus PS = – 12

, dan

PS melalui titik (–4, 0). Jadi, persamaan bagi garis lurus PS ialah y2 – y1 = m(x2 – x1)

y – 0 = – 12

(x + 4)

y = – 12

x – 2

12 (a) Min anggaran jisim bagi guli-guli ini = (5 × 5 + 10 × 16 + 15 × 20 + 20 × 24 +

25 × 18 + 30 × 10 + 35 × 7) ÷ 100 = 1 960 ÷ 100 = 19.6 g(b)

Jisim KekerapanKekerapan longgokan

3 – 7 5 5

8 – 12 16 21

13 – 17 20 41

18 – 22 24 65

23 – 27 18 83

28 – 32 10 93

33 – 37 7 100

(c)

2.5 0

20

10

30

40

50

71

7.5 12.5 17.5 22.5 27.5 32.5 37.5

60

70

80

90

100

Kekerapan longgokan

Jisim (g)

25

(d) Bilangan guli dengan jisim sekurang-kurangnya 25 g = 100 – 71 = 29

13 (a) Palsu (Sebilangan persamaan kuadratik tidak mempunyai punca)

(b) Pernyataan 1: 8 ialah nombor genap. Pernyataan 2: 8 ialah gandaan 2.(c) Premis 2: 4 bukan faktor bagi 30.

14 (a) (i) 145° 34′ terletak di sukuan II dan kos 145° 34′ adalah negatif.

Jadi kos 145° 34′ = –kos (180° – 145° 34′)

= –kos 34° 26′ = –0.8248

(ii) 218° 25′ terletak di sukuan III dan tan 218° 25′ adalah positif.

Jadi, tan 218° 25′ = tan (218° 25′ – 180°)

= tan 38° 25′ = 0.7931(b) sin θ = –0.7234. Sudut tirus bagi θ = 46° 20′. Jadi θ = 180° + 46° 20′ = 226° 20′.(c) (i) y = kos x (ii) p = 360° – 64° = 296°

15 (a)

Markah Kekerapan Titik tengah

21 – 30 2 25.5

31 – 40 4 35.5

41 – 50 9 45.5

51 – 60 5 55.5

61 – 70 5 65.5

71 – 80 2 75.5

81 – 90 3 85.5

(b) 10(c) Min anggaran markah = (25.5 × 2 + 35.5 × 4 + 45.5 × 9 +

55.5 × 5 + 65.5 × 5 + 75.5 × 2 + 85.5 × 3) ÷ 30

= 1 615 ÷ 30 = 53.83 (d)

Keke

rapa

n

2

4

5

3

1

Markah25.5 35.5 45.5 55.5 65.5 75.5 85.5

6

7

8

9

10

0

(e) Selang kelas mod = 41 – 50

16 (a) Persamaan bagi garis lurus AB ialah y = 2.

(b) Kecerunan AD = 4 – 20 – 2

= –1

Pintasan-y = 4. Jadi persamaan garis lurus

AD ialah y = –x + 4.(c) Kecerunan BC = kecerunan AD = –1 Katakan C = (0, y).

Jadi, y – 20 – 5

= –1

y = 7 Pintasan-y bagi garis lurus BC ialah 7. (d) Pintasan-y = 7 (0, 7), 3(7) + 7(0) = k k = 21

3y + 7x = 21 (x, 0), 3(0) + 7x = 21 x = 2 Titik persilangan garis lurus dengan

paksi-x ialah (3, 0).

Soalan KBAT

Bab 1 Luas taman = 20.25 m2

Oleh itu, panjang tepi taman = √20.25 = 4.5 mPanjang tepi luar pejalan kaki itu = 4.5 m + 1.5 m + 1.5 m = 7.5 mMaka, jumlah luas pejalan kaki itu= 7.52 – 20.25 = 36 m2

Jumlah kos membina laluan pejalan kaki itu = RM53 × 36 = RM1 908 = RM1 900

Bab 2x = 100

x – 15x2 – 15x = 100(x – 20)(x + 5) = 0x = 20 @ x = –5 (ditolak)Purata kelajuan = 20 km j–1

Bab 3(a) 25 (b) 75(c) 50

Bab 4(a) n2 + 1, n = 1, 2, 3, 4, … (b) 170

Bab 70.15 × 900 = 135

Bab 860°

Bab 10Bangunan A = (60 × sin 40°) + 15 = 53.567 m

Bab 11(a) APD atau DPA (ditanda pada rajah)(b) tan PQR = 6

24

PQR = 14.04˚ atau 14˚ 2′

Membaca

75 5025

Aktiviti luar

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