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Matematik Tingkatan 3 Bab 1
8. (a) Padankan setiap berikut dengan bentuk (b) Isikan petak kosong di ruang jawapan
termudah. dengan nombor yang betul.
Match each of the following with its simplest form. Fill in the boxes in the answer space with correct numbers.
[3 markah/ 3 marks] [2 markah / 2 marks]
Jawapan/ Answer : Jawapan/ Answer :
1 2
12
3
2
2
(p × p ) p –2 8 = (2 ) = (2 )
2
3 2
8 = (8) 8 = (2 )
2
2
2
3 2
3 3
p 2 = (2 ) = 2
6
2
1 2
12
p
4
1 2 2 p 2 = (2 )
p
5
Bahagian C
p 10
10. (a) Selesaikan setiap persamaan yang berikut.
4
(√p ) p 12 Solve each of the following equations.
3 2
[1 markah/ 1 mark]
(i) 3 × 3 = 3
2
4
x
(ii) 2 × 4 = 32 [2 markah/ 2 marks]
y
2
(b) Tandakan (✓) jika pernyataan di ruang Jawapan/ Answer :
2
4
x
jawapan adalah benar dan (7) jika tidak. (i) 3 × 3 = 3
2 + x
4
Mark (✓) if the statement in the answer space is true and 3 = 3
(7) if not. 2 + x = 4
[1 markah/ 1 mark] x = 2
Jawapan/ Answer : (ii) 2 × 4 = 32
2
y
5
Nombor dalam tatatanda indeks boleh 2 × (2 ) = 2
y
2 2
ditulis sebagai a dengan a dikenali 2 × 2 = 2 5
4
y
n
5
sebagai indeks. 2 y + 4 = 2
A number in index notation can be written as a where y + 4 = 5
n
a is known as index. y = 1
7
(b) Selesaikan persamaan serentak yang berikut.
a , a dikenali sebagai asas dan n dikenali Solve the following simultaneous equations.
n
x
y
sebagai indeks. 2 × 2 = 128
a is known as the base and n is known as the index. 6 x = 36 y
6
[4 markah/ 4 marks]
9. (a) Rajah di bawah menunjukkan satu Jawapan/ Answer :
persamaan. 2 × 2 = 128
y
x
The diagram below shows an equation. 2 x + y = 2
7
p × p × q 4 = p q x + y = 7 …… 1
2
3
n –2
q m 6 x
= 36 y
6
Nyatakan nilai m dan nilai n. 6 x – 1 = (6 )
2 y
State the value of m and of n. x – 1 2y
[2 markah/ 2 marks] 6 = 6
Jawapan/ Answer : x – 1 = 2y
p × p × q 4 x – 2y = 1 …… 2
2
3
= p q
n –2
q m 1 – 2: y – (–2y) = 7 – 1
q
p 3 + 2 4 – m = p q 3y = 6
n –2
p q = p q y = 2
n –2
5 4 – m
Gantikan y = 2 ke dalam 1:
n = 5, 4 – m = –2 Substitute y = 2 into 1:
m = 6 x + 2 = 7
x = 5
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