03[a math cd]
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1 © Penerbitan Pelangi Sdn. Bhd.
Paper 1
1. (e3)2 = e3 × 2
= e6
Answer: D
2. (h1—2 )8 = h
1—2
× 8
= h4
Answer: B
3. 5−3 = 153
Answer: B
4. t−
4—3 = 1
t4—3
= 1––––3ABt4
Answer: D
5. (9n4)1—2 = (9)
1—2 (n4)
1—2
= AB9 (n4 ×
1—2 )
= 3n2
Answer: C
6. (8x3)2—3 = (8)
2—3 (x3)
2—3
= (3AB8 )2(x3 ×
2—3 )
= (2)2(x2) = 4x2
Answer: D
7. p4
–––9
3—2
= (p4)
3—2
–––––(9)
3—2
= p
4 × 3—2
––––––(AB9 )3
= p6
––––(3)3
= p6
–––27
Answer: D
8. (m2)4 × m−2 = m2 × 4 × m−2
= m8 + (−2)
= m6
Answer: A
9. (p3)4 ÷ p5 = p3 × 4 ÷ p5
= p12 − 5
= p7
Answer: C
10. (x4)2 = x2n
x4 × 2 = x2n
x8 = x2n
∴ 8 = 2n n = 4
Answer: B
11. 3−2 × 27 = 3m
3−2 × 33 = 3m
3−2 + 3 = 3m
31 = 3m
∴ m = 1
Answer: C
CHAPTER
3 IndicesCHAPTER
2
Mathematics SPM Chapter 3
© Penerbitan Pelangi Sdn. Bhd.
12. (2q3)2 × 3q−4 = (2q3)(2q3) × 3q−4
= 2 × 2 × 3 × q3 + 3 + (−4)
= 12q2
Answer: C
13. (2e2)3 ÷ 4e = (2)3(e2 × 3) ÷ 4e
= 8e6–––4e1
= 2e6 − 1
= 2e5
Answer: D
14. (5−1 × 31—2 )2 = (5−1)2 × (3
1—2
× 2)
= 5−2 × 31
= 1–––52
× 3
= 3–––25
Answer: C
Paper 1
1. 1t n = 6
–
1—2
t –n = 6–
1—2
∴ t = 6, n = 12
Answer: B
2. (h4k3)–1
h2k –5
= h– 4k –3
h2k –5
= h– 4 − 2k −3 − (−5)
= h–6k2
Answer: D
3. 203—5 = x 20y
203—5 = 20
y—x
∴ x = 5, y = 3
Answer: A
4. (16m4)1—4
––––––––2n
× m3n4
= 24 ×
1—4 m
4 ×
1—4
2n
× m3n4
= 2m2n
× m3n4
= m1 + 3 × n4 – 1
= m4n3
Answer: C
5. p3
4q3 = p3q–3
4
Answer: C
6. n × n
5—4
–––––n
1—4
= n1—2 × n
5—4 × n
–
1—4
= n1—2
+
5—4
+ (–
1—4
)
= n3—2
Answer: C
7. 1 23 2–5
= 1 23 2–5 ×
1—2
= 1 23 2
–
5—2
= 1 32 2
5—2
Answer: C
8. (2a3 × 3b5—2 ) 2–––––––––––
a2b3
= 22a3 × 2 × 32b5—2 ×
2
––––––––––––––a2b3
= 4a6 × 9b5
a2b3
= 36a6 – 2b5 – 3
= 36a4b2
Answer: C
3
Mathematics SPM Chapter 3
© Penerbitan Pelangi Sdn. Bhd.
9. 1 23 2–3
= 1
1 23 23
= 1827
= 1 ÷ 827
= 1 × 278
= 278
Answer: B
10. 2x = 16–––23x
2x = 24–––23x
2x = 24 − 3x
∴ x = 4 − 3x x + 3x = 4 4x = 4 x = 1
Answer: B
11. 6–
1—2
= 1––––6
1—2
Answer: D
12. 42 × 31—2
–––––––62
4
= 42 × 4 × 31—2
× 4
–––––––––––62 × 4
= 48 × 32––––––
68
= (22)8 × 32––––––––(2 × 3)8
= 216 × 32–––––––28 × 38
= 216 − 8 × 32 − 8
= 28 × 3−6
Answer: B
Paper 1
1. h5—4 × (h
1—4 × k
4—3 )3
= h5—4 × (h
1—4
× 3k
4—3
× 3)
= h5—4 × h
3—4 k4
= h5—4
+
3—4 × k4
= h2k4
Answer: D
2. (2t−3m)2 × 5t2
= (22t−3 × 2m1 × 2) × 5t2
= 4t−6m2 × 5t2
= 20t−6 + 2m2
= 20t−4m2
Answer: A
3. (3h−2k)3 × 1—3 h2k−1
= (33h−2 × 3k1 × 3) × 3−1h2k−1
= 33 + (−1)h−6k3 × h2k−1
= 32h−6 + 2k3 + (−1)
= 9h−4k2
Answer: D
4. 2m3n––––––––
(3mn−2)2
= 2m3n–––––––––––32m1 × 2n−2 × 2
= 2m3n––––––9m2n−4
= 2m3 − 2n1 − (−4)–––––––––––
9
= 2mn5–––––
9Answer: C
5. (ef 4)2 ÷ e−3f 9= (e1 × 2f 4 × 2) ÷ e−3f 9= e2f 8 ÷ e−3f 9= e2 − (−3)f 8 − 9
= e5f −1
Answer: A
4
Mathematics SPM Chapter 3
© Penerbitan Pelangi Sdn. Bhd.
6. (x−2y3)3 × x7y2
= (x−2 × 3y3 × 3) × x7y2
= x−6y9 × x7y2
= x−6 + 7y9 + 2
= xy11
Answer: C
7. (2p2q1—3 )3 × p3q−4
= (23p2 × 3q1—3
× 3) × p3q−4
= 8p6q1 × p3q−4
= 8p6 + 3q1 + (−4)
= 8p9q−3
Answer: B
8. (27m3n9)1—3 ÷ 9m2n−1
= (271—3 m
3 ×
1—3 n
9 ×
1—3 ) ÷ 9m2n−1
= 3m1n3 ÷ 9m2n−1
= 3m1 − 2n3 − (−1)
–––––––––––9
= m−1n4–––––
3Answer: C
9. 3g × (16h6)1—2
–––––––––––(g9h–3)
1—3
= 3g × 16
1—2 h
6 × 1—2
––––––––––––g
9 × 1—3 h
−3 ×
1—3
= 3g × 4h3––––––––
g3h–1
= 12g1 − 3h3 − (−1)
= 12g−2h4
= 12h4–––––
g2
Answer: D
10. (x3y1—3 )2 ÷ (x6y5)
1—3
= (x3 × 2y1—3
× 2) ÷ (x
6 ×
1—3 y
5 ×
1—3 )
= x6y2—3 ÷ x2y
5—3
= x6 − 2y2—3
−
5—3
= x4y−1
Answer: C
11. [2p × (3p)1—2 ]4 ÷ (p–12)
2—3
= [(2p)4 × (3p)2] ÷ p–8
= (2p)4 × (3p)2
p–8
= 144p14
Answer: B
12. (16 × 3−4)1—2 ÷ (23 × 3−5)
= (161—2 × 3
−4 ×
1—2 ) ÷ (23 × 3−5)
= (22 × 3−2) ÷ (23 × 3−5)= 22 − 3 × 3−2 − (−5)
= 2−1 × 33
= 33
21
= 272
Answer: A
13. (27 × 2−6)1—3 × (2−4 × 3)
= (271—3 × 2
−6 ×
1—3 ) × 2−4 × 31
= 31 × 2−2 × 2−4 × 31
= 31 + 1 × 2−2 + (−4)
= 32 × 2−6
= 32–––26
= 9–––64
Answer: B
14. (41—3 × 5)3 ÷ 56
–––24
= (41—3
× 3 × 53) ÷ 56
–––24
= (22 × 53) ÷ (56 × 2−4)= 22 − (−4) × 53 − 6
= 26 × 5−3
= 26–––53
= 64––––125
Answer: C
5
Mathematics SPM Chapter 3
© Penerbitan Pelangi Sdn. Bhd.
15. (35—2 × 54)
1—2 × 3
3—4 × 5−3
= (35—2
×
1—2 × 5
4 ×
1—2 ) × 3
3—4 × 5−3
= 35—4 × 52 × 3
3—4 × 5−3
= 35—4
+
3—4 × 52 + (−3)
= 32 × 5−1
= 32–––51
= 9––5
Answer: A
16. (9 × 2−6)1—2 ÷ (2−5 × 34)
= (91—2 × 2
−6 ×
1—2 ) ÷ (2−5 × 34)
= (31 × 2−3) ÷ (2−5 × 34)= 31 − 4 × 2−3 − (−5)
= 3−3 × 22
= 22–––33
= 4–––27
Answer: C
17. (641—3 × 5−1)2 ÷ (5−4 × 23)
= (4 × 5−1)2 ÷ (5−4 × 23)= (22 × 2 × 5−1 × 2) ÷ (5−4 × 23)= (24 × 5−2) ÷ (5−4 × 23)= 24 − 3 × 5−2 − (−4)
= 21 × 52
= 2 × 25= 50Answer: A
18. 813—4 × (32)3
––––––––––96
= (34)
3—4 × 32 × 3
–––––––––––(32)6
= 33 × 36––––––
312
= 33 + 6 − 12
= 3−3
= 1–––33
= 1–––27
Answer: C
19. 125−
2—3 × (34)
1—2
––––––––––––25–2
= (53)−
2—3 × 3
4 ×
1—2
––––––––––––(52)−2
= 5−2 × 32–––––––
5−4
= 5−2 − (−4) × 32
= 52 × 32
= 25 × 9= 225
Answer: D
20. 324—5
–––––––––––(212 × 7−4)
1—2
= (25)4—5
––––––––––––2
12 ×
1—2 × 7
−4 ×
1—2
= 24–––––––26 × 7−2
= 24 − 6 × 72
= 2−2 × 72
= 72–––22
= 49–––4
Answer: C
21. 27−
5—3
––––––––––9−4 × (4
1—3 )6
= (33)−
5—3
––––––––––––(32)−4 × 4
1—3
× 6
= 3−5–––––––3−8 × 42
= 3−5 − (−8) × 4−2
= 33 × 4−2
= 33–––42
= 27–––16
Answer: B
6
Mathematics SPM Chapter 3
© Penerbitan Pelangi Sdn. Bhd.
22. 8–––27
4—3 ÷ (5
−
1—2 × 4
1—2 )4
= 23–––33
4—3 ÷ (5
−
1—2
× 4 × 4
1—2
× 4)
= 2—3
4 ÷ (5−2 × 42)
= 24–––34
÷ (5−2 × 24)
= 24 – 4––––
34 × 52
= 20×52
––––––34
= 25–––81
Answer: A
23. (36 × 7−4)1—3 ÷ (7
2—3 × 9−1)
= (36 ×
1—3 × 7
−4 ×
1—3 ) ÷ (7
2—3 × 3−2)
= (32 × 7−
4—3 ) ÷ (7
2—3 × 3−2)
= 32 − (−2) × 7− 4—
3 −
2—3
= 34 × 7−2
= 34–––72
= 8149
Answer: D
24. 2p – 1 + 2p + 1 – 2p
= 2p(2 –1 + 2 – 1)
= 2p 12
+ 1
= 2p × 32
= 3(2p – 1)
Answer: D
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