Page 13 - PRAKTIS HEBAT MATHS TG 2

Matematik Tingkatan 2 Jawapan

4. (a) (i) 3 (ii) 28 unit 2
Prakt is PT 3 Koordinat (b) (10, –8)
7
(c) (8, 6), (–8, 6), (–8, –6), (8, –6)
Coordinates
KBAT
A B
1. B 2. A 3. C 4. D 5. B
6. B 7. C 8. B 9. B 10. C
Prakt is PT 3 Graf Fungsi
8
B Graph of Functions
1. Pasangan titik Jarak (Unit) A
Pair of points Distance (Units)
3 1. C 2. B 3. A 4. C 5. D
A dan / and B 6. B 7. D 8. C 9. B 10. C
4
5 B
A dan / and C
6
1.
7 Hubungan satu
C dan / and D kepada satu
8 One-to-one relation
9
B dan / and E Hubungan satu
10 kepada banyak
One-to-many relation
2. (a) 3 (b) 3 (c) 3 (d) 3
Hubungan banyak
3. AB = [6 – ( 0 )] – [4 – ( 4 )] 2 kepada satu
2
Many-to-one relation
2
= 6 – 0 2
= 6 unit Hubungan banyak
kepada banyak
4. Many-to-many relation
(0, 0) (2, 1) (–2, 4)
2. (a) BENAR / TRUE
(3, 9) (3, –1) (8, 9) (b) PALSU / FALSE
(c) BENAR / TRUE
(d) PALSU / FALSE
C 3. (a) 2, 3, 4, 5, 6 (c) {10, 12}
(b) 10, 12 (d) {2, 3, 4, 5, 6}
1. (a) Q(10, –4) 4. (a) ✗ (b) 3
(b) (12, 5), (12, –5), (–12, 5) dan / and (–12, –5) (c) 3 (d) ✗
(c) (i) S (ii) 37.5 unit 2
C

2. (a) (i) E 6, 7  (ii) 3 unit / 3 units 1. (a) a = 3, b = 16
2
(b) (i) 5.39 unit / 5.39 units
(ii) –1 (b) (i) x –3 –2 –1 0 1 2
(c) P(5, 14)
3. (a) (i) E (ii) (–6, –1) y 12 6 2 0 0 2
(b) A(2, 9)
(c) (i) Q (ii) x y
(ii) P dan Q / P dan R / Q dan R / S dan Q.
Kerana titik tengah antara titik-titik tersebut –3
berada di dalam bulatan. Bangku tersebut tidak –2 0
boleh berada di dalam kolam. –1 2
P and Q / P and R / Q and R / S and Q. 0 6
Because the midpoints between these points 1 12
are located in the circle. The bench cannot be 2
placed in the pool.
(iii) SR = 12.17 unit / units Jenis hubungan: Fungsi banyak kepada satu
PS = 12 unit / units Type of relation: Many-to-one function

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15 JAW HEBAT Mate Tg2.indd 5 27/04/2020 10:50 AM