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PRAKTIS BAB 1Soalan Objektif 1 C 2 C 3 B 4 D 5 A 6 D 7 A 8 A 9 D 10 A11 D 12 C 13 B 14 A 15 C16 C
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(d) x –3 –2 –1 0 1 2
y 59 21 7 5 3 –11
y
x1 2 3–1–2–3 0
50
60
30
20
10
–10
40
y = –2x3 + 5
(e) x –4 –3 –2 –1 1 2 3 4
y –7.5 –10 –15 30 30 15 10 7.5
y
x
y = 30
x
1 2 3–1–2–3 0 4–4
25
20
10
5
–5
–10
–15
–20
–25
–30
30
15
(f) x –4 –3 –2 –1 1 2 3 4
y 5.5 7.3 11 22 –22 –11 –7.3 –5.5
y
x
y = –22
x
1 2 3–1–2–3 0 4–4
25
20
10
5
–5
–10
–15
–20
–25
15
2 (a) (i) y = –14 (ii) x = –1.8, 2.3 (b) (i) y = 10 (ii) x = –1.5 (c) (i) y = 6 (ii) x = 1.8
3 (a) (c)
y
x10
45
y
x0
6
3
(b) (d)
y
x0
5
5
y
–3
10x
4 (a) y
x
5
2 5–1 0
x = 2 (b) y
x–1
–3
0
x =
14
14
32
(c)
y
x
5
–1 0
x =
34
34
52
5 (a) y
x0–3
–27
(b)
y
x0–2
24
6 (a) (b) y
x0
y
x0
(c)
y
x0
Latihan Bestari 2.1 1 (a) y = 2x2 + 1 Apabila x = –2, y = 2(–2)2 + 1 = 9 Apabila x = 2, y = 2(2)2 + 1 = 9
x –3 –2 –1 0 1 2 3
y 19 9 3 1 3 9 19
(b)
y
x
20
15
10
5
–5
0
6.5
y = 2x2 + 1
1 2 3–1–2–3–1.6 1.65 2.6
(c) (i) Apabila x = 2.6, y = 14. (ii) Apabila y = 6.5, x = –1.6 dan 1.65.
2 Untuk mencari pintasan-x, y = 0. x2 – 8x – 9 = 0 (x + 1)(x – 9) = 0 x = –1, x = 9 Untuk mencari pintasan-y, x = 0. [ y = 0 – 0 – 9 = –9 Untuk mencari paksi simetri,
x = – b2a
y
x–1 90–9
x = 4
4 = – –8
2(1) = 4
3 y + 1 = x3
y = x3 – 1 Untuk mencari pintasan-x, y = 0. [ x3 – 1 = 0 x = 3√1 = 1 Untuk mencari pintasan-y, x = 0. [ y = –1
y
x0
–11
4 2x + y = 5 Pintasan-x, y = 0. [ 2x + 0 = 5
x = 52
Untuk mencari pintasan-y, x = 0. [ y = 5
x
5
y
0 52
2.2 Penyelesaian Persamaan dengan Kaedah Graf 1 (a)
25
20
15
–5
1 2 3
y
x
y = x2 + 5x + 1
–4 0(0.8, 6)
y = –5x + 10
–1–2–3
10
5
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(b)
25
20
10
5
1 2 3–1–2–3
y
0
–5
–10
–15
–20
–25
x4–4
15
(1.1, –20)
y = –5x – 15
y = – 22x
(c)
20
30
10
1 2
y
y = 3x3 – 1
–40
–50
–60
–70
–80
(–2.8, –72)
y = 15x – 30
–10
–20
–30
–1–2–3 0x
–90
2 (a)
–5
1 2 3–1–2–3
y
x
y = 3x2 – 2x – 1
0
35
(1.75, 4.5)(–0.8, 2.2)
30
25
20
15
10
5y = x + 3
(b)
60
50
30
20
10
–1–2–3
y
y = –2x3 + 5
–10
–20
–30
–40
x
40
(–2.7, 47)
y = –10x + 20
0 1 2 3
↓
Latihan Bestari 2.2 1 (a) Graf fungsi: y = 2x2 + 3x – 1 ⇒ 2x2 = y – 3x + 1 Gantikan 2x2 = y – 3x + 1 ke dalam
2x2 + x – 6 = 0 y – 3x + 1 + x – 6 = 0 y = 2x + 5 (b) Graf fungsi: y = –3x2 – x + 7 ⇒ 3x2 = –y – x + 7
⇒ x2 = 13
(–y – x + 7)
Gantikan x2 = (–y – x + 7) ke dalam x2 + x + 5 = 0. x2 + x + 5 = 0 1
3(–y – x + 7) + x + 5 = 0
–y – x + 7 + 3x + 15 = 0 –y + 2x + 22 = 0 y = 2x + 22 (c) Graf fungsi: y = –2x3 + 5 ⇒ 2x3 = 5 – y Gantikan 2x3 = 5 – y ke dalam
2x3 + x + 7 = 0. 2x3 + x + 7 = 0 5 – y + x + 7 = 0 y = x + 12 (d) Graf fungsi: y = 3x3 + x – 7 3x3 = y – x + 7
x3 = 13
(y – x + 7)
Gantikan x3 = 13
(y – x + 7) ke dalam x3 – x + 1 = 0. x3 – x + 1 = 0
13
(y – x + 7) – x + 1 = 0
y – x + 7 – 3x + 3 = 0 y = 4x – 10 (e) 2x2 + 5x = 16
2x + 5 = 16x
2x + 5 = y
[ y = 2x + 5
2.3 Rantau yang Mewakili Ketaksamaan dalam Dua Pemboleh Ubah
1 (a) y
x0
y = –x + 2
(b) y
x0
y = 1
2x – 1
(c) (d)
y
x0
y =
2x +
4
y
x0
y =
5x –
6
2 (a) y x = 6
x + y = 6
0
y = x + 412
x
(b)
y
x0
y = x
+ 5
y = – 12 x
(c)
y
x0
y = 2x
– 6
y = –x + 2
y = 1
(d)
y
x0
y = 3x + 5
y = 5y = x
3 (a) x < 0 (b) x . –4 y > –x y >
12 x
y , 6 x + y < 0
(c) y < 6x x + y < 7 3y – x . 0 4 x < 0, y > –x –5 dan y < x + 4.
Latihan Bestari 2.3 1 y = 2x – 3 Apabila x = 0, y = 0 – 3 = –3. y
x
–3
10
–1
Apabila x = 1, y = 2 – 3 = –1. Pilih titik (0, 0). 0 2(0) – 3 0 > –3 Maka, titik (0, 0) memuaskan y > 2x – 3. 2 y = –x + 6 Apabila x = 0, y = 0 + 6 = 6. Apabila y = 0, x = 6. Pilih titik (0, 0). 0 0 + 6
y
x
6
0 6 0 , 6 Maka, titik (0, 0) memuaskan y , –x + 6. 3 Gantikan x = 2 dan y = 4 ke dalam ketaksamaan y . 2x – 9. y . 2x – 9 4 . 2(2) – 9 4 . 4 – 9 4 . –5 (Benar) Titik A(2, 4) berada dalam kawasan y . 2x – 9.
Anjakan Prima Math F5 Jaw 3rd.indd 3 9/10/2017 3:52:54 PM
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4 Gantikan x = –4 dan y = 0 ke dalam ketaksamaan y . 4x + 1. y . 4x + 1 0 . 4(–4) + 1 0 . –16 + 1 0 . –15 (Benar) Titik B(–4, 0) berada dalam kawasan y . 4x + 1.
PRAKTIS BAB 2Soalan Objektif 1 C 2 B 3 D 4 A 5 D 6 C 7 A 8 B 9 B 10 A11 D 12 C 13 A 14 B 15 B16 C 17 D 18 B 19 D 20 C
Soalan Subjektif
1 (a) Apabila x = –3, y = – 6(–3)
= 2
Apabila x = 0.5, y = – 60.5
= –12
(b)y
x
y = – 6x
1 2 3–1–2–3 0–4 4
12
10
6
4
2
–2
–4
–6
–8
–10
8
–12
y = –x – 2
(c) (i) y = –3.8 (ii) x = –0.9
(d) 6x = x + 2
⇒ –y = x + 2 y = –x – 2 Daripada graf, x = –3.6 dan 1.7. 2 (a) y = 2x2 – 3x – 2 k = 2(–3)2 – 3(–3) – 2 = 25 m = 2(2)2 – 3(2) – 2 = 0
(b) y
x1 2 3–1–2–3 0 4–4
45
40
30
25
20
10
5
35
–5
15
y = –2x + 28
y = 2x2 – 3x – 2
(c) (i) y = 18 (ii) x = –1.7, 3.2 (d) y = 2x2 – 3x – 2 ⇒ 2x2 = y + 3x + 2 Gantikan persamaan ke dalam 2x2 – x – 30 = 0. y + 3x + 2 – x – 30 = 0 y = –2x + 28 Daripada graf, x = –3.65. 3 (a) Apabila x = –2, y = (–2)2 – (–2) – 2 = 4 Apabila x = 0.5, y = (0.5)2 – (0.5) – 2 = –2.25
(b)y
x1 2 3–1–2–3 0 4
12
10
6
4
2
–2
–4
8
y = –2x + 8 y = x2 – x – 2
(c) (i) y = 3 (ii) x = –2.5, 3.45 (d) y = x2 – x – 2 ⇒ x2 = y + x + 2 Gantikan persamaan ke dalam x2 + x – 10 = 0. y + x + 2 + x – 10 = 0 y = –2x + 8 Daripada graf, x = 2.7.
BAB 3 Penjelmaan III
3.1 Gabungan Dua Penjelmaan
1 (a) 6
y
x
y = –1
AA�
C�
C��B��
A��
B�
B C
R
T4
2
–2
–4
–6
20–2–4–6 4 6
(b)
6
y
x
4
2
–2
–4
–6
20–2–4–6 4 6
B�
C�
y = –1
x = 1
A�
A�
B�
C�
BC
A
(c)
6
y
x
4
2
–2
–4
–6
20–2–4–6 4 6C��
B��
A��
D��
C�
D�
A�
B�
A
B
CD
y = –x
Anjakan Prima Math F5 Jaw 3rd.indd 4 9/10/2017 3:52:55 PM
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(c) y
x
4
2
–2
–4
20–2–4–6–8 4 6C��
A��
B�
C�
A�
B
A
C
x = –1
(d) y
x
4
2
–2
–4
20–2 4 6 8
C��
B�� A��
C�
A�
B�AB C
(e) y
x
4
2
–2
–4
20–2–4–6–8 4 6A��
C��
B��
A� B�
C�
A B
C
(f)
6
y
x
4
2
–2
–4
420–2–4–6 6 8A��
B��
y =
x
C��C�B�
A�
BA
C
4 (a) (2, 4) (e) (3, 4) (b) (–6, –1) (f) (4, 2) (c) (6, –4) (g) (–2, 4) (d) (–4, –2) 5 (a) PQ tidak sama dengan QP (b) PQ tidak sama dengan QP
6 (a) (i) U ialah translasi 8–1 .
(ii) V ialah pantulan pada garis y = –1.
(b) (i) U ialah translasi –1 5 .
(ii) V ialah putaran 90° ikut arah lawan jam melalui (0, 0).
(c) (i) U ialah translasi 74 .
(ii) V ialah pembesaran dengan faktor skala 2 pada pusat (1, 1).
(d) (i) U ialah pantulan pada garis x = –1.
(ii) V ialah translasi –7–4 .
(e) (i) Q ialah pantulan pada garis x = –1. (ii) P ialah putaran 90° ikut arah jam
melalui (0, 0). 7 (a) Putaran 180° ikut arah jam melalui (0, 0). (b) Putaran 180° ikut arah jam melalui (0, 0). (c) Putaran 90° ikut arah lawan jam melalui (2, 4).
Latihan Bestari 3.1 1
6
y
x
A�A�� 4
2
–2–2 0–4 2 4
A
Koordinat imej = (–4, 4)
2
6
y
x
P��4
2
–2–2 0–4 2 4
P�P
Koordinat imej = (1, 5)
3
y
PQ
RC
B
A
x
8
6
4
2
2
(7, 9)
04 6 8 10 12
V ialah putaran 90° ikut arah jam melalui (7, 9).
SUDUT KBAT 1 (a) N adalah pembesaran pada pusat T
dengan faktor skala 12
. M adalah pantulan
pada garis US.
(b) QSU = 30 × 12 2
QSU = 7.5 cm2
Luas kawasan berlorek = 30 – 7.5 = 22.5 cm2
PRAKTIS BAB 3Soalan Objektif 1 C 2 B 3 C 4 D 5 B 6 C 7 A 8 C 9 B 10 C11 B 12 C 13 C 14 C 15 D16 D
Soalan Subjektif 1
P�
P��
P
10
y
x
y = 5
8
6
4
2
20
4 6 8 10 12
(d)
6
y
x
4
2
–2
–4
–6
20–2–4–6 4 6C��
B��
A��
C�
B�
A�
B
AC
(e)
6
y
x
4
2
–2
–4
–6
20–2–4–6 4 6
C�� A��
B��
A�
B�
C�
B
A
C
(f)
6
y
x
4
2
–2
–4
–6
20–2–4–6 4 6
C��
B��
A��
C�
B�
A�
B
C
A
2 (a) II (c) I (b) II (d) II 3 (a)
6
y
x
4
2
–2
–4
–6
20–2–4–6–8 4 6B
C
A
A��
C��
B��
C�
A� B�
(b)
6
y
x
4
2
–2
–4
–6
20–2–4–6–8 4 6BA
C
C�
A� B�
C��
A�� B��
Anjakan Prima Math F5 Jaw 3rd.indd 5 9/10/2017 3:52:56 PM
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(a) A : Pantulan pada garis y = 5
(b) B : Translasi 54
2 y
x
x = 6
8
6
4
2
20
4 6 8 10 12 14
A�B�/B��
C�
A��
C��
A B
C
(a) Q : Pantulan pada garis x = 6 (b) P : Pembesaran dengan faktor skala 2 pada pusat (8, 3). 3 10
y
x
8
6
4
2
20
4 6 8
(5, 2)
90°
10 12 14
P��
P�P
(a) A : Putaran 90° ikut arah lawan jam (6, 5) (b) B : Pembesaran dengan faktor skala 3 pada pusat (5, 2). 4
y
x
8
6
4
2
2
(1, 1)0
4 6 8 10 12 14 16 18
P��
P�
P
(a) A : Pembesaran dengan faktor skala 2 pada pusat (1, 1).
(b) B : Translasi 7–2 .
5 Luas objek = 11 cm2
Faktor skala = 2 [ Luas imej = 22 3 11 cm2
= 4 3 11 cm2
= 44 cm2
6 Luas objek = 18 cm2
Faktor skala = 2 Luas imej = 22 3 18 cm2
= 72 cm2
[ Luas kawasan berlorek = 72 – 18 = 54 cm2
7 (a)
8
y
x
y = 5
6
4
2
0–2 42 8 10
(8, 7)
(6, 4)
(8, 5)
6 (i) (8, 7) (ii) (8, 5)
(b)
y
C BF G
E AD
N M
x
y = x + 1
K L
8
6
4
2
20
4 6 8 10 12 14 16
(i) V ialah pantulan pada garis y = x + 1. (ii) Putaran 90° ikut arah lawan jam melalui titik (6, 7). (c) (i)
10
y
x
N M
K Z
YX
L
8
6
4
2
20
4 6 8 10 12 14
(7, 1)
16
Pusat pembesaran = (7, 1) (ii) Luas imej = 530 cm2
Faktor skala = 2 Luas objek
= 530 cm2
(2)2 = 132.5 cm2
8 (a)
8
10
y
x
y = 3
6
4
2
–20–2 42 8 10
(8, 7)
(8, –1)
(5, 1)
6
(i) (8, –1) (ii) (8, 7)
(b) y
x
B C
A D
F
HE
G
8
6
4
2
20
4 6 8 10 12 14
U ialah putaran 90° ikut arah lawan jam melalui titik (6, 7). W ialah pembesaran dengan faktor
skala 2 pada pusat (7, 6). (c) Andaikan ABC = x cm2. Luas imej = 22 3 x 20 + x = 4x 3x = 20
x = 203
= 6 23
cm2
9 (a)
4
y
x
2
–4
–2
–2 0–4 2 4 6
M
M�
N��
N�
N2�
N2��
N
(i) Imej of M = (–2, 4)
(ii) (a) Imej of N = (–4, 2) (b) Imej of N = (1, 0) (b)
6
y
A
B
DC
FE
GH
M
L
J
K
x = –1
x
4
2
2–4 –2 0–6 2 4 6
(3, 4)
8
(i) (a) V ialah pantulan pada garis x = –1. (b) W ialah pembesaran dengan faktor skala 3 pada pusat (3, 4). (ii) Luas imej = 167.4 cm2
Faktor skala = 3 Luas objek = 167.4 cm2
32
= 18.6 cm2
[ Luas kawasan berlorek = 167.4 cm2 – 18.6 cm2
= 148.8 cm2
10 (a)
6
y
x
4
2
–2–2 0–4 2 4 6
(3, 0)
(0, 2)
(–2, –1)
(i) (–2, –1) (ii) (0, 2) (iii) (3, 0) (b)
y
G E D A B
C
O
F
x
x = 9
6
4
2
2 4 6 8 10 12 14 (i) U ialah pantulan pada garis x = 9. (ii) V ialah pembesaran dengan faktor skala 2 pada pusat (8, 6). (c) Luas objek = 16 cm2
Faktor skala = 2 Luas imej = 22 3 16 cm2
= 64 cm2
[ Luas kawasan berlorek = 64 cm2 – 16 cm2 = 48 cm2
Anjakan Prima Math F5 Jaw 3rd.indd 6 9/10/2017 3:52:57 PM
4.8 Penyelesaian Persamaan Linear Serentak dengan Kaedah Matriks
1 (a) 2 –13 1 x
y = –1 6
(b) 3 –1–2 1 x
y = 5–3
(c) 2 00 3 x
y = 74
2 (a) 2 –1–1 3 x
y = 7–11
xy =
1(6 – 1)
3 11 2 7
–11
= 15 21 – 11
7 – 22
= 2–3
\ x = 2, y = –3
(b) 2 11 –3 x
y = 11–5
xy =
1(–6 – 1)
–3 –1–1 2 11
–5
= – 17 –33 + 5
–11 – 10
= 43
\ x = 4, y = 3
(c) 5 22 –1 x
y = –4–7
xy =
1(–5 – 4) –1 –2
–2 5 –4–7
= – 19 4 + 14
8 – 35
= –23
\ x = –2, y = 3
(d) 5 23 –1 x
y = –7–13
xy =
1(–5 – 6)
–1 –2–3 5 –7
–13
= – 111
7 + 2621 – 65
= –34
\ x = –3, y = 4
(e) 2 33 –1 x
y = –11–11
xy =
1(–2 – 9)
–1 –3–3 2 –11
–11
= – 111
11 + 3333 – 22 = –4
–1 \ x = –4, y = –1
Latihan Bestari 4.8
1 2 –13 1 x
y = 61
2 –1 5 6 –3 x
y = 2–5
3 2 1–3 2 x
y = –511
xy = 1
4 + 3 2 –13 2 –5
11 = 1
7 –10 – 11–15 + 22
= –3 1
[ x = –3, y = 1
4 4 –3–3 1 x
y = 5–5
xy = 1
4 – 9 1 33 4 5
–5 = 1
–5 5 – 1515 – 20
= 21
[ x = 2, y = 1
SUDUT KBAT 1 (a) P + Q = 230 8P + 3Q = 1 290
1 18 3 P
Q = 2301 290
(b) 1 18 3 P
Q = 2301 290
PQ = 1 1
8 3 –1
2301 290
= 1(1)(3) – (1)(8)
3 –1–8 1 230
1 290 = – 1
5 (3)(230) + (–1)(1 290)(–8)(230) + (1)(1 290)
= – 15 –600
–550 P
Q = 120110
P = 120 Q = 110 P – Q = 120 – 110 = 10
Anjakan Prima Math F5 Jaw 3rd.indd 8 9/10/2017 3:52:58 PM
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PRAKTIS BAB 4Soalan Objektif 1 A 2 C 3 B 4 B 5 D 6 B 7 A 8 C 9 D 10 C11 D 12 B 13 D 14 A 15 A16 A
Soalan Subjektif
1 (a) M = 12 + 15 1 3
–5 2 =
117
317
– 5
17
217
(b) 2 –3
5 1 xy = –7
8 x
y = 1
17
317
– 5
17
217
–7 8
= 13
[ x = 1, y = 3
2 (a) 2 31 –2
–1 = 1
–4 – 3 –2 –3–1 2
= – 17 –2 –3
–1 2 = m –2 p
–1 2
[ m = –17 , p = –3
(b) 2 31 –2 x
y = –1–4
xy = – 1
7 –2 –3–1 2 –1
–4 = – 1
7 2 + 121 – 8
= –2 1
[ x = –2, y = 1
3 (a) Q = P–1
= 1
4 + 3 2 3–1 2
= 17 2 3
–1 2 = k 2 h
–1 2 [ k =
17 , h = 3
(b) 2 –31 2 x
y = –110
xy = 1
7 2 3–1 2 –1
10 = 1
7 –2 + 301 + 20
= 43
[ x = 4, y = 3
4 (a) 3 1–1 2
–1 = 1
6 + 1 2 –11 3
= 27
– 17
17
37
= 27
– 17
17
h [ h = 3
7
(b) 3 1–1 2 x
y = 5–4
xy = 1
7 2 –11 3 5
–4 = 1
7 10 + 45 – 12= 2
–1 [ x = 2, y = –1
5 (a) Q = 1–3 – 8 –3 –4
–2 1 =
311
411
2
11
– 111
(b) 1 4
2 –3 xy = 1
13
xy = – 1
11 –3 –4–2 1 1
13 = – 1
11 –3 – 52–2 + 13
= 5–1
[ x = 5, y = –1
6 (a) 1 24 5
–1 = 1
5 – 8 5 –2–4 1
= – 13 5 –2
–4 1 = n 5 q
–4 1
[ n = –13 , q = –2
(b) 1 24 5 u
v = –1–7
uv = – 1
3 5 –2–4 1 –1
–7 = – 1
3 –5 + 144 – 7
= – 13 9
–3 = –3 1
[ u = –3, v = 1
7 (a) R = P –1
= 1–2 – 3 –1 –3
–1 2
= – 15 –1 –3
–1 2 = k –1 h–1 2
[ k = –15 , h = –3
(b) 2 31 –1 u
v = –6 7
uv = – 1
5 –1 –3–1 2 –6
7 = – 1
5 6 – 216 + 14
= – 15 –15
20 = 3–4
[ u = 3, v = –4
8 (a) 4 –32 1
–1 = 1
4 + 6 1 3–2 4
= 110 1 3
–2 4
= 1
10
310
– 1
5
25
[ p =
25
(b) 4 –32 1 u
v = 1–7
uv = 1
10 1 3–2 4 1
–7 = 1
10 1 – 21–2 – 28
= –2–3
[ u = –2, v = –3
9 (a) R = 1–2 – 3 –2 –1
–3 1
= – 15 –2 –1
–3 1 =
25
15
35
– 15
(b) 1 13 –2 u
v = –4–7
uv = – 1
5 –2 –1–3 1 –4
–7 = – 1
5 8 + 712 – 7
= –3–1
[ u = –3, v = –1
10 (a) 1 15 –2
–1 = 1
–2 – 5 –2 –1–5 1
= – 17 –2 –1
–5 1 = p –2 –1
–5 q [ p = –
17 , q = 1
(b) 1 15 –2 x
y = 1–16
xy = – 1
7 –2 –1–5 1 1
–16
Anjakan Prima Math F5 Jaw 3rd.indd 9 9/10/2017 3:52:59 PM
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= – 17 –2 + 16
–5 – 16 = – 1
7 14–21 = –2
3
[ x = –2, y = 3
BAB 5 Ubahan
5.1 Ubahan Langsung
1 (a) Ya (b) Tidak (c) Ya (d) Tidak (e) Tidak 2 (a) y = kx 2 = k(8) k = 1
4
y = x4
(b) k = 51
y = 5√x
(c) k = 510
= 12
y = √x2
(d) k = 5025
= 2
y = 2x2
(e) k = 168
= 2
y = 2x
3 (a) k = 305
= 6
y = 6x
y = 6(7) = 42
(b) k = 510
= 1
2
n = 12
(7)
= 3 . 5
(c) y = kx2
43
= k(4)
k = 1
3
y = 13
(25)
= 25
3
(d) y = kx2
92
= k(9)
k = 1
2
w = 12
(16)
= 8
(e) y = kx3
16 = k(8) k = 2 y = 2(125) = 250
(f) p = k√ t 5 = k√100
k = 1
2
3 = 12
√n
n = 62
= 36
(g) p = kq2
43
= k(4)
k = 1
3
3 = 13
q2
q2 = 9
q = 3
(h) u = k√ v 2 = k√100
k = 1
5
5 = 15
√ n
√ n = 25 n = 625
Latihan Bestari 5.1 1 p x2
⇒ p = kx2
p1
x12 =
p2
x22
2422
= p42
244
= p
16
p = 96
2 y x3
⇒ y = kx3
y1
x13 =
y2
x23
4523
=
100x3
x3 = 1 000 x = 10
3 p √q ⇒ p = k√q p1
√q1
= p2
√q2
1√49
= w√81
w = 97
5.2 Ubahan Songsang 1 (a) Ya (b) Tidak (c) Ya
(d) Tidak (e) Tidak 2 (a) k = 128 p = 128
q2
(b) k = 30 × 2 = 60 y = 60
√ x (c) k = 21 × 4 = 84 y = 84
√ x (d) k = –36 × 2 = –72 p = – 72
q
(e) k = –648 P = – 648
q3 3 (a) k = 9 × 4 = 36 y = 36
72
y = 1
2
(b) k = 72 × 4 = 288 y = 288
9 = 32
(c) k = 18 × 16 = 288 p = 288
36
= 8 (d) k = 80 × 4 = 320 y = 320
10
= 32 (e) k = 80 × 1
8
= 10 640 = 10
q3
q3 = 1
64
q = 1
4
Latihan Bestari 5.2
1 y = 1x2
⇒ y = kx2
x12 y1 = x2
2 y2
4218 = 82y
y = 2
64
= 1
32
Anjakan Prima Math F5 Jaw 3rd.indd 10 9/10/2017 3:52:59 PM
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2 y 1x3
⇒ y = kx3
x13 y1 = x2
3 y2 23(54) = x3(16)
x3 = 43216
= 27
x = 3
3 p 1√q
⇒ p = k√q
p1√q1 = p2√q2
n√64 = 16
(√144)
8n = 2
n = 14
5.3 Ubahan Tercantum
1 (a) y rt
(b) p 1qt
(c) p q2
t 2 (a) y = k
r2t
– 1
6 = k4(3)
6k = –12 k = –2 y = –2
r2t
(b) y = kq2r
k = –279
= –3 y = –3q2r
(c) y = k√ rt
1 =
k√ 48
2k = 8 k = 4 y = 4√ r
t
(d) p = kt√ r
– 12 = k
–4 × 4
2k = 16 k = 8 p = 8
t√ r
(e) p = kqx3
– 5 = k
1 × 8
4 4k = –40 k = –10 p = –10
qx3
(f) y = kr3
p –2 = k(–8)
2 –4 = –8k k =
12
y = r3
2p 3 (a) –1 = k(–2)
5
2k = 5 k = 5
2
n =
52 (4)
5 = 2 (b) k = –36
6 = –6 y = –6pq 30 = –6(–4)(w) 24w = 30 w = 5
4
(c) 1 = k(4)8
4k = 8 k = 2 6 = 2(9)
w 6w = 18 w = 3 (d) 1
2 =
k(3)18
6k = 18 k = 3 3 =
3(10)n
3n = 30 n = 10 (e) –8 = k12 (8) 4k = –8 k = –2 2 = –2(w)(4) –8w = 2 w = – 1
4
(f) –
=
1
2 k
112
(–8)
k = 13
w = 1
319
(27)
= 1
9
Latihan Bestari 5.3
1 y √rq
y = k√rq
2 = k√166
k = 3
[ y = 3√r
q
2 y p2
q y = kp2
q 1
3 = k 3 22
36 k = 3
[ y = 3p2
q
3 p
1qx3
p = kqx3
23
= k1
24 3 23
k = 29
[ p = 2
9qx3
4 p 1rq
p = krq
p1 q1 r1 = p2 q2 r2
–2 23
3 3 3 14
= 112
3 6 3 q
q = –4
5 y √rp
y1p1
√r1
= y2p2
√r2
(–2) 3 10√100
= 3 3 4√r1
√r = –6 r = 36
SUDUT KBAT
1 Masa a KeluasanPekerja
2 = k(200)8
k = 0.08
(a) t = 0.08(400)8
t = 4 hari (b) 4 = 0.08(L)
10
L = 500 Luas = 500 m2
PRAKTIS BAB 5Soalan Objektif 1 B 2 A 3 D 4 B 5 A 6 B 7 C 8 D 9 C 10 C11 D 12 B 13 C 14 C 15 C16 A 17 B 18 B 19 A 20 A
Anjakan Prima Math F5 Jaw 3rd.indd 11 9/10/2017 3:52:59 PM
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BAB 6 Kecerunan dan Luas di bawah Graf
6.1 Kuantiti yang diwakili oleh Kecerunan Graf 1 (a) Kadar perubahan jarak melawan masa
atau laju dalam m s–1. (b) Kadar perubahan laju melawan masa atau
pecutan dalam km j–2.
2 (a) Jarak (km)
Masa(jam)
24
1620
84
12
10
2 3
(b) Jarak (m)
Masa (s)
6
45
21
3
10
2 3 3 (a) Kecerunan OP = 3 m s–1
Kecerunan PQ = 5 m s–1
Objek bergerak dengan laju seragam 3 m s–1 pada lapan saat pertama. Kemudian, ia menambah kelajuan kepada 5 m s–1 dan 7 saat lagi diperlukan untuk objek tiba di titik B. (b) Kecerunan OP = 80 km j–1
Kecerunan PQ = 0 Kecerunan QR = –64 km j–1
Bas itu bergerak dari Bandar A ke Bandar B dengan laju seragam 80 km j–1.
Kemudian, bas itu berhenti selama setengah jam. Selepas itu, bas itu bergerak
semula dari B ke A dengan laju seragam 64 km j–1.
4 (a) Laju = 124
= 3 m s–1
(b) Laju = 20 – 86
= 2 m s–1
(c) 8 – 2 = 6 saat (d) 0 m s–1 (kerana zarah ketika itu adalah
pegun) (e) Laju = 18
6
= 3 m s–1
(f) Laju = 248
= 3 m s–1
5 (a)
0g
2
N(RM)
2
12
Kadar satu pusingan ialah RM5. Caj permulaan ialah RM2.
(b) Isi padu (cm3)
0 10
80
Masa (s)
Isipadu (cm3)
Air mengalir pada kadar 8 cm3 sesaat.
Latihan Bestari 6.1 1 Kuantiti yang diwakili oleh kecerunan ialah
faedah, dalam RM, yang dikumpulkan daripada akaun deposit tetap yang selari dengan masa dalam tahun.
2 Kecerunan OP = 52 – 01 – 0
= 52 km j–1
Kecerunan PQ = 52 – 521.5 – 1
= 0 km j–1
Kecerunan QR = 0 – 522 – 1.5
= –104 km j–1
Bas itu bergerak dari bandar X ke bandar Y dengan laju purata 52 km j–1. Kemudian bas itu berhenti selama setengah jam. Kemudian, bas itu bergerak dari bandar Y ke bandar X dengan laju seragam 104 km j–1.
3 (a) Laju = Kecerunan
= 32 – 205 – 0
= 2.4 m s–1
(b) Laju = Kecerunan
= 0 – 3217 – 11
= –5 13
m s–1
4 (a) Tempoh = 12 s – 5 s = 7 saat (b) Laju = Kecerunan
= 0 – 2017 – 12
= –4 m s–1
(Laju negatif menandakan bahawa objek itu bergerak ke titik rujukan.)
6.2 Kuantiti yang diwakili oleh Luas di bawah Graf 1 (a) Harga yang perlu dibayar oleh pelanggan bagi mangga yang dibeli. (b) Jarak yang dilalui oleh zarah yang bergerak. 2 (a) Luas A + Luas B = 1
2 × (15 + 8) × 6 + 1
2 × 15 × 4
= 23 × 3 + 15 × 2 = (69 + 30) unit2
= 99 unit2
(b) Luas A + Luas B + Luas C = 1
2 × 8 × 10 + 10 × 7 + 1
2 × (20 + 10) × 6
= 40 + 70 + 90 = 200 unit2
3 (a) Jarak = 12
× 18 × 34 = 306 m
(b) Jarak = 12
× 25 × 8 = 100 m (c) Jarak = 30 × 10 = 300 m
(d) Jumlah jarak = 1
2 × (30 + 18) × 10 + (14 × 18)
= 240 + 252 = 492 m
(e) Jarak = (15 × 10) + 12
× (35 + 15) × 5 = 150 + 25 × 5 = 275 m (f) Jumlah jarak = 36 × 20 + 1
2 (a) Tempoh masa = 1.0 jam – 0.6 jam = 0.4 jam (b) Purata laju
= Jumlah jarak yang dilaluiJumlah masa yang diambil
= 102 km3 jam
= 34 km j–1
(c) Graf ABCD menyilang graf AE pada ketika 0.9 jam dan jaraknya ialah 60 km.
(i) Maka, kereta dan bas berjumpa pada jarak 60 km. (ii) Bas itu memerlukan 0.9 jam untuk tiba di lokasi. 3 (a) Tempoh masa = 15 s – 6 s = 9 saat (b) Kadar perubahan laju = Kecerunan
= 12 – 86 – 0
= 46
= 23
m s–2
(c) Jumlah jarak yang dilalui = 222 1
2 3 (8 + 12) 3 6 +
[(15 – 6) 3 12] + = 222 1
2 3 (12 + 24) 3 (t – 15)
60 + 108 + (18t – 270) = 222 18t = 324 t = 18 s
4 (a) Laju seragam = 18 m s–1
(b) (i) Jarak yang dilalui = 144 (t – 16) 3 18 = 144 t – 16 = 8 t = 24 s (ii) Purata laju
= Jumlah jarak yang dilaluiJumlah masa yang diambil
12
3 (24 + 18) 3 16 + [144] +
=
12
3 (30 – 24) 3 18 30 = 336 + 144 + 54
30 = 17.8 m s–1
5 (a) Tempoh masa = 20 s – 12 s = 8 saat (b) Kadar perubahan laju = Kecerunan
= 46 – 1012 – 0
= 3 m s–2
(c) Jumlah jarak yang dilalui = 796 m 1
2 3 (10 + 46) 3 12 +
[8 3 46] + = 796
12
3 (t – 20) 3 46 336 + 368 + 23t – 460 = 796 23t = 552 t = 24 s 6 (a) Laju seragam = 20 m s–1
(b) (i) Jumlah jarak = 160 (t – 11) 3 20 = 160 t – 11 = 8 t = 19 s (ii) Purata laju
= Jumlah jarak yang dilaluiJumlah masa yang diambil
12
3 (8 + 19) 3 20 +
=
12
3 (20 + 35) 3 6 25 = 270 + 165
25
= 17.4 m s–1
7 (a) Tempoh masa = 18 s – 13 s = 5 saat (b) Kadar perubahan laju = Kecerunan
= 0 – 2825 – 18
= –4 m s–2
(c) Jumlah jarak yang dilalui = 400 m
12
3 (v + 28) 3 13 + [(18 – 13) 3 28]
= 400
132
v + 182 + 140 = 400
v = 78 3 213
= 12 m s–1
Anjakan Prima Math F5 Jaw 3rd.indd 13 9/10/2017 3:53:00 PM
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BAB 7 Kebarangkalian
7.1 Kebarangkalian Suatu Peristiwa 1 (a) S = {kepala, ekor} (b) S = {7, 8, 9, 10} (c) S = {k, k, b, b, b, b} (d) S = {P, Q, R, S, T} (e) S = {m, b, b, u, u, u} (f) S = {1, 1, 5, 5, 5, 5, 5, 10}
2 (a) 12
(d) 15
(b) 16
(e) 15
(c) 25
(f) 25
3 (a) 13
= 21
21 + x
21 + x = 63 x = 63 – 21 = 42 biji
(b) 12
= 1016 + x
16 + x = 20 x = 4
(c) 310
= 1218 + x
54 + 3x = 120 3x = 66 x = 22
(d) 25
× 20 = 8
(e) 12
= 1527 + x
27 + x = 30 x = 3
(f) 37
= 2130 + x
90 + 3x = 147 3x = 57 x = 19
(g) 13
= 1319 + x
19 + x = 39 x = 20
(h) 516
= 4065 + x
325 + 5x = 640 5x = 315 x = 63
Latihan Bestari 7.1 1 (a) {10, 11, 12, 13, 14, 15, 16, 17, 18, 19} (b) Nombor perdana = 11, 13, 17, 19
P(Nombor perdana) = 410
= 25
2 (a) {C, H, O, C, O, L, A, T, E} (b) Vowels = O, O, A, E
P(Vowels) = 49
3 P(durian) = 1812 + 18 + x
13
= 1812 + 18 + x
30 + x = 54
x = 24
7.2 Kebarangkalian Pelengkap Suatu Peristiwa 1 (a) (i) T9 ialah peristiwa mendapat nombor yang lebih besar daripada 2. (ii) T9 = {3, 4, 5, 6} (b) (i) P9 ialah peristiwa mendapat nombor yang bukan nombor perdana. (ii) P9 = {1, 4, 6} (c) (i) G9 ialah peristiwa memilih pelajar
lelaki. (ii) G9 = {lelaki} (d) (i) V9 ialah peristiwa memilih kad huruf konsonan. (ii) V9 = {B, C, D, F, G, H, J} (e) (i) A9 ialah peristiwa memilih oren. (ii) A9 = {oren}
2 (a) 1 – 47
= 37
(b) 1 – 38
= 58
(c) 1 – 710
= 310
(d) 1 – 45 = 1
5
(e) 1 – 5
9 = 49
Latihan Bestari 7.2 1 (a) Andaikan B ialah peristiwa memilih sebiji guli biru. [ B9 ialah peristiwa memilih guli hijau. (b) P(B9) = 1 – P(B)
= 1 – 411
= 7
11 2 (a) Andaikan A ialah peristiwa mengambil nombor perdana. [ A9 ialah peristiwa mengambil nombor bukan nombor perdana. (b) A9 = 1 – P(A)
= 1 – 29
= 7
9
7.3 Kebarangkalian Peristiwa Bergabung
1 (a) (i) A B = {2, 3, 4, 6} (ii) A B = {6} (b) (i) A B = {1, 2, 3, 5} (ii) A B = {3, 5}
2 (a) 59
(c) 136
(b) 14
8 (a) Tempoh masa = 1.8 jam – 0.9 jam = 0.9 jam (b) Purata laju = 140 km
2.5 jam = 56 km j–1
(c) Graf ABCD dan AEF bersilang pada jarak 110 km. Maka, teksi dan bas berjumpa pada jarak 110 km dari bandar R. 9 (a) Laju seragam = 12 m s–1
(b) (i) Jarak yang dilalui = 108 m (t – 8) 3 12 = 108 t – 8 = 9 t = 17 s (ii) Purata laju
= Jumlah jarak yang dilaluiJumlah masa yang diambil
12
3 (20 + 12) 3 8 + 108 +
=
12
3 (12 + 18) 3 (20 – 17) 20
= 128 + 108 + 4520
= 14.05 m s–1
10 (a) Tempoh masa = 18 s – 6 s = 12 saat (b) Kadar perubahan laju = Kecerunan
= 15 – 06 – 0
= 2.5 m s–2
(c) Jumlah jarak yang dilalui = 295 m 1
2 3 (12 + 18) 3 15 +
= 295 1
2 3 (15 + v) 3 (22 – 18)
225 + 30 + 2v = 295 2v = 40 v = 20 m s–1
11 (a) Laju seragam = 16 m s–1
(b) (i) Jumlah jarak dilalui = 208 (28 – t) 3 16 = 208 28 – t = 13 t = 15 s (ii) Purata laju
12
3 (10 + 20) 3 9 +
= 1
2 3 (20 + 16) 3 (15 – 9)
+ 208 28
= 135 + 108 + 20828
= 16.1071
= 16.11 m s–1
Anjakan Prima Math F5 Jaw 3rd.indd 14 9/10/2017 3:53:00 PM
4 (a) Bearing B dari A ialah 040°. (b) Bearing A dari B ialah 105°. (c) Bearing P dari Q ialah 210°. (d) Bearing Q dari P ialah 320°. (e) Bearing B dari A ialah 065°. (f) Bearing D dari C ialah 108°. (g) Bearing E dari F ialah 242.7°. 5 (a) 020° (b) 116° (c) 260° (d) 225°
(e) 318° 6 (a) 260° (b) 260°
(c) 320°
Latihan Bestari 8.1 1
U
Q
P
230°
2 90° + 24° = 114° Bearing Q dari P ialah 114°. 3
U U
Q
P115°
65° 65°
115°
Bearing Q dari P ialah 115°.
BAB 8 Bearing
8.1 Bearing 1 (a) (i) Utara (ii) Timur (iii) Barat (b) (i) Timur laut (ii) Tenggara (iii) Selatan (iv) Barat daya (v) Barat (vi) Barat laut 2 (a) 075° (b) 140° (c) 230° (d) 220° (e) 295° 3 (a)
U A
B
70°
(b) U
A
B
110°
Anjakan Prima Math F5 Jaw 3rd.indd 16 9/10/2017 3:53:02 PM
= 3 076.819 = 3 076.8 batu nautika 4 Jarak AB = Beza longitud 3 60 3 kos (latitud) 490 = θ 3 60 3 kos 41° 329
θ = 10° 559
[ Longitud B = (21° 179 + 10° 559) B = (31° 729) B= 32° 129B 5 Jarak terpendek antara P dan Q = (180° – 50° – 50°) 3 60 = 4 800 batu nautika
SUDUT KBAT 1 Daripada rajah, P = (80° 179U, 20°B) dan Q = (17° 269U, 20°B). Jarak PQ = (80° 179 – 17° 269) 3 60 = (62° 5193 60) = 3 771 batu nautika
Masa yang diambil = JarakLaju
= 3 771450
= 8.38 jam
= 8 jam 23 minit Waktu berlepas = (2210 – 0823) jam = Jam 1347
PRAKTIS BAB 9Soalan Objektif 1 C 2 B 3 B 4 B 5 C 6 C 7 D 8 A 9 A 10 B11 D 12 A 13 A 14 A 15 A16 B 17 A 18 C 19 A 20 D
Soalan Subjektif 1 (a) Longitud = (180° – 15°) B = 165°B
(b)
15°T165°B
U
42°U
42°S
S
OP
R
R ialah (42°S, 165°B). (c) Jarak terpendek = (90° – 42°) 3 60 = 2 880 batu nautika (d) (i) Jarak PM = Laju 3 masa = 480 3 6 = 2 880 batu nautika (ii) Jarak PM = θ 3 60 3 kos 42° 2 880 = θ 3 60 3 kos 42° θ = 64° 359
Longitud M = (64° 359 – 15°) B = 49° 359B 2 (a) Longitud Q = (180° – 60°) B = 120°B (b) Jarak PR = θ 3 60 5 400 = θ 3 60 θ = 90° Latitud R = (90° – 40°) U = 50°U (c) Jarak PQ = Beza longitud 3 60 3 kos 40° = 180° 3 60 3 kos 40° = 8 273.2799 = 8 273.3 batu nautika (d) Jarak Q ke P = (90° – 40°) 3 2 3 60 = 50° 3 2 3 60 = 6 000 batu nautika Jarak P ke R = 5 400 batu nautika
Masa yang diambil = JarakLaju
= 6 000 + 5 400600
= 19 jam
3 (a)
20°T160°B
U
30°S
30°U
S
OP
J
JP ialah diameter bumi, kedudukan P ialah (30°S, 160°B).
(b) (i) Jarak JL = Beza longitud 3 60 3 kos 30°
θ = 76° 599
y = 76° 599 – 20° = 56° 599
(ii) Jarak JK = Beza latitud 3 60 3 600 = a 3 60
a = 60° [ x = 60° – 30°
= 30°
Anjakan Prima Math F5 Jaw 3rd.indd 18 9/10/2017 3:53:03 PM
J19
(c) Jarak JL = 4 000 batu nautika Jarak LM = (30° + 40°) 3 60 = 4 200 batu nautika Masa yang diambil
= JarakLaju
= 4 000 + 4 200540
= 15 jam 11 minit
4 (a) Kedudukan Q = (40°U, 30°B) Longitud R = (180° – 30°) T = 150°T Kedudukan R ialah (40°U, 150°T). (b) Jarak terpendek = Beza longitud 3 60 = (90° – 40°) 3 60 3 2 = 6 000 batu nautika (c) Jarak PX = θ 3 60 6 200 = θ 3 60 θ= 103° 209
[ Latitud X = (103° 209 – 40°) S = 63° 209S (d) Jarak PR = Beza longitud 3 60 3 kos 40° = 150° 3 60 3 kos 40° = 6 894.4 batu nautika Masa yang diambil
= JarakLaju
= 6 894.4540
= 12 jam 46 minit
5 (a) PQ ialah diameter selarian latitud. Longitud Q = (180° – 20°) B
= 160°B (b)
20°T
58°S
58°U
160°B
U
S
O
R
P
PR ialah diameter bumi. R ialah (58°S, 160°B). (c) Jarak terpendek = (90° – 58°) 3 60 = 1 920 batu nautika (d) (i) Jarak PM = Purata laju 3 Masa yang diambil = 600 3 8 = 4 800 batu nautika (ii) Jarak PM = Beza longitud 3 60 3 kos 58° 4 800 = θ 3 60 3kos 58° θ = 150° 589
[ Longitud M = (20° + 150° 589) T = 170° 589T 6 (a) Jarak PR = Beza latitud 3 60 3 600 = θ 3 60 θ= 60°
[ Latitud R = (60° – 50°) U = 10°U (b) Jarak PQ = Beza longitud 3 60 3 kos 50° = 180° 3 60 3 kos 50° = 6 942.1 batu nautika (c) PQ ialah diameter selarian latitud. Longitud Q = (180° – 40°) B = 140°B (d) Jarak QP = (90° – 50°) 3 2 3 60 = 4 800 batu nautika Jarak PR = 3 600 batu nautika Masa yang diambil
= DistanceSpeed
= 4 800 + 3 600500
= 16 jam 48 minit
7 (a)
140°T40°B
U
30°S
30°U
S
O
A
C
AC ialah diameter bumi. Maka, kedudukan C ialah (30°U, 140°T). (b) Jarak AB = (70° – 40°) 3 60 3 kos 30° = 1 558.8 batu nautika (c) Jarak AQ = θ 3 60 3 kos 30° 5 000 = θ 3 60 3 kos 30° θ= 96° 139
[ Longitud Q = (96° 139 – 40°) T = 56° 139T 8 (a) (i) Longitud R = (180° – 30°) B = 150°B (ii) Jarak PQ = (40° + 30°) 3 60 3 kos 21° = 3 921 batu nautika
(b) Jarak QX = θ 3 60 4 800 = θ 3 60 θ= 80° [ Latitud X = (80° – 21°) U = 59°U 9 (a) (i) 30°S. (ii) Longitud P = (180° – 24°) T = 156°T (b) Jarak PQ = 180° 3 60 3 kos 30° = 9 353.1 batu nautika (c) Jarak terpendek PQ = (180° – 30° – 30°) 3 60 = 7 200 batu nautika
Masa yang diambil = JarakLaju
= 7 200450
= 16 jam
Masa = 0230 + 1600 = 1830 jam 10 (a) Latitud Q ialah 35°S. (b) Jarak = (90° – 35°) 3 60 = 3 300 batu nautika (c) Jarak PX = θ 3 60 3 000 = θ 3 60 θ= 50° [ Latitud X = (50° – 35°) S = 15°S (d) Jarak PY = (70° – 42°) 3 60 3 kos 35° = 1 376.2 batu nautika
Masa yang diambil = JarakLaju
= 1 376.2320
= 4 jam 18 minit
BAB 10 Pelan dan Dongakan
10.1 Unjuran Ortogon 1 (a)
QP
RS
(b)
QP
RS
(c)
QP
RS
(d)
QP
RS
(e)
PQ
RS
Anjakan Prima Math F5 Jaw 3rd.indd 19 9/10/2017 3:53:04 PM
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Latihan Bestari 10.1 1
QP
RS
2
C
A B
E
D
PQ
SR
10.2 Pelan dan Dongakan 1 (a)
F/BE/A 5 cm
G/CH/D
3 cm
(b)
E
A
F/B
H
D
G/C
3 cm
2 cm
6 cm
(c)
L/G/B
M/H/C
E/A
J/D
4 cm
K/F
N/I
4 cm2 cm
(d)
H/B
I/C
E/A
L/D
4 cm
F G
K J
2 cm3 cm 1 cm (e)
G/B
H/C
E/A
J/D
4 cm
F
I
2 cm 3 cm
2 (a) (i)
B/C
V
5 cm
6 cm
A/D
(ii)
6 cm
C/D
V
4 cmB/A
(b) (i)
B/C
G/H
L/M
A/D
E/J
K/N
F/I
3 cm
3 cm
4 cm
6 cm
(ii)
C/D
M/N
H/I/J
B/A
L/K
G/F/E
3 cm
4 cm
3 cm
(c) (i)
B/C
V
5 cm
4 cm
A/D
(ii)
C/D
V
4 cm
4 cm
B/A
(d) (i)
B/C
H/IG/J
6 cmA/D
F/K3 cm
1 cm
2 cm3 cm
E/L
(ii)
A/B4 cmD/C
E/F
G/HJ/I
L/K
2 cm3 cm
1 cm
(e) (i)
B/C
G/H
3 cm
A/D
E/JF/I2 cm
4 cm
5 cm
(ii)
A/B4 cmD/C
E/F
G1 cm
3 cm
H
J/I
3 (a)
Q10 cm
7.5 cm
P
RT/S
(b)
B/C14 cmA/D
F/G4 cm
15 cm
E/H
(c)
M/N
U
10 cm
2 cm
5 cm
S/T
L/K
P
R/Q
SUDUT KBAT 1 (a)
B
E
F
DC
5 cm
6 cm 10 cm
A
(b)
B
E F
J I
H
C
G
D
6 cm
5 cm
3 cm
2 cm
10 cm
A
Latihan Bestari 10.2
1 (a)
F/E
G/H
L/B
M/C
K/A
N/D
3 cm 3 cm
4 cm
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(b)
E/H
F/G
3 cm 3 cm
2 cm
B/C
L/M
3 cm
A/D
K/N
5 cm
(c)
A/E/B
F
L
K
D/H/C
G
M
N
2 cm
4 cm
2 cm
1 cm
PRAKTIS BAB 10Soalan Subjektif 1 (a)
E/L
F/K
3 cm
4 cm
2 cmI/A
J/B
H/D
G/C
(b) (i)
A/B/P
I/JL/K
E/F
M/N
D/C/Q
H/G/R
4 cm
2 cm
2 cm
2 cm
5 cm
6 cm
(ii)
P/Q
N/RM/G
B/C4 cm 2 cmA/D
I/L
H
E
J/K
F
6 cm
2 cm
2 cm
2 cm
2 (a)
G/HE/JF/I
A/BD/C 5 cm
2 cm2 cm
1.5 cm3 cm
(b) (i)
G/AE/D
J/C IH/B P/K
S/NR/M Q/L
F 3 cm
3 cm2 cm
2 cm2 cm
2cm
(ii)
L/MA/D
G/F/E
H/S/I/J
K/B/N/C
R
QP
2 cm
2 cm3 cm
4 cm
3 (a)
B/C
G/H
J/I
A/D
F/E
6 cm
2 cm
4 cm4 cm
(b) (i)
K/A
Q/M
P/N
G/L
J/B
E/D
H
I/C2 cm
4 cm
2 cm3 cm
(ii)
A/BD/C 5 cm
1 cm
2 cmK/L
P/Q
N/M
E/H
2 cm
2 cm
G
JI
4 (a)
B/C
F
G
A/D
E
H2 cm
4 cm
5 cm
(b) (i)
C/D
G/H
6 cm
1 cm
J/B/A
F/E
4 cm4 cm
5 cm L/K
M/N
(ii)
J
M/LN/K
5 cm
E/A
H/D G/C
6 cm
1 cm
5 cm F/B
5 (a)
5 cm
4 cm2 cm
B/C
I/JE/H
F/G
A/D
(b) (i)
N/K
M/L
E/A
H/DG Q/J/C
F
I/P/B
2.5cm
2.5cm
2cm
6 cm
(ii)
2 cm
2 cm
2 cm
G
J/H2 cm
1 cm FN/P
I/E
M/Q
1 cm6 cm L/C/DK/B/A
6 (a)
4 cm
5 cm B/CA/D2 cm
2 cmH
F1 cm
G
E
(b) (i)
H/BS/E/A 5 cm
2cm
F/C
6 cm
R/M
Q/L P/N/K
G/D
4 cm
(ii)
C/D
K/L
P/Q
FN
G
4 cm
4 cm
4 cm
2 cmB/A/M
H
E
S/R
2 cm
2 cm
1 cm1 cm
7 (a)
B/C
G
F
2 cm
4 cmA/D
H
E
2 cm
3 cm
(b) (i)
E/A
H/D
F/B
K/G/C
L/M/N
6 cm
3 cm
4 cm
1cm
(ii)
B/A
F/E
5 cm
G/M/H
C/N/D
K/L
6 cm
2 cm
6 cm1.5cm
8 (a)
B/C
E
F
3 cm
2 cm
5 cm
3 cm
3 cm3 cm
3 cm
A/D
I
J
G
H
3 cm
2 cm
(b) (i)
E/B
F/C
I/A
J/D
6 cm
2.5cm
2.5cm
H
Anjakan Prima Math F5 Jaw 3rd.indd 21 9/10/2017 3:53:09 PM
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(ii)
C/D
F/JH
E/I
B/A 6 cm
5 cm3 cm
Kertas Model SPM
KERTAS 1 1 A 2 D 3 A 4 A 5 B 6 A 7 C 8 D 9 D 10 B11 B 12 B 13 D 14 C 15 C16 A 17 B 18 A 19 B 20 C21 D 22 B 23 B 24 A 25 B26 A 27 B 28 B 29 D 30 C31 C 32 A 33 A 34 D 35 A36 B 37 A 38 B 39 C 40 D