mediastreet.com.my_jawapan/impak a+ maths tg 4...mediastreet.com.my
TRANSCRIPT
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 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
J2 Global Mediastreet Sdn. Bhd. (762284-U)
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
J3 Global Mediastreet Sdn. Bhd. (762284-U)
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
J4 Global Mediastreet Sdn. Bhd. (762284-U)
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
�
Anjakan Prima Math F4 Jaw 4th.indd 4 9/10/2017 3:50:31 PM
J5 Global Mediastreet Sdn. Bhd. (762284-U)
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.
Anjakan Prima Math F4 Jaw 4th.indd 5 9/10/2017 3:50:31 PM
J6 Global Mediastreet Sdn. Bhd. (762284-U)
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
Anjakan Prima Math F4 Jaw 4th.indd 6 9/10/2017 3:50:31 PM
J7 Global Mediastreet Sdn. Bhd. (762284-U)
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
Anjakan Prima Math F4 Jaw 4th.indd 7 9/10/2017 3:50:32 PM
J8 Global Mediastreet Sdn. Bhd. (762284-U)
(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
Anjakan Prima Math F4 Jaw 4th.indd 8 9/10/2017 3:50:32 PM
J9 Global Mediastreet Sdn. Bhd. (762284-U)
(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
Anjakan Prima Math F4 Jaw 4th.indd 9 9/10/2017 3:50:33 PM
J10 Global Mediastreet Sdn. Bhd. (762284-U)
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
Anjakan Prima Math F4 Jaw 4th.indd 10 9/10/2017 3:50:34 PM
J11 Global Mediastreet Sdn. Bhd. (762284-U)
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
Anjakan Prima Math F4 Jaw 4th.indd 11 9/10/2017 3:50:34 PM
J12 Global Mediastreet Sdn. Bhd. (762284-U)
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°
Anjakan Prima Math F4 Jaw 4th.indd 12 9/10/2017 3:50:35 PM
J13 Global Mediastreet Sdn. Bhd. (762284-U)
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°
Anjakan Prima Math F4 Jaw 4th.indd 13 9/10/2017 3:50:35 PM
J14 Global Mediastreet Sdn. Bhd. (762284-U)
(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
Anjakan Prima Math F4 Jaw 4th.indd 14 9/10/2017 3:50:36 PM
J15 Global Mediastreet Sdn. Bhd. (762284-U)
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′
Anjakan Prima Math F4 Jaw 4th.indd 15 9/10/2017 3:50:36 PM
J16 Global Mediastreet Sdn. Bhd. (762284-U)
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
Anjakan Prima Math F4 Jaw 4th.indd 16 9/10/2017 3:50:37 PM