Vol 1 Chapter 1
-
Upload
jm-garrick -
Category
Documents
-
view
96 -
download
2
Transcript of Vol 1 Chapter 1
1
Exam Practice 1
1. x + 2 – x – 6 = x – 3 Squaring both sides, (x + 2) + (x – 6) – 2( x + 2)( x – 6) = x – 3 x – 1 = 2( x + 2)( x – 6) Squaring both sides, x2 – 2x + 1 = 4(x2 – 4x – 12) x2 – 2x + 1 = 4x2 – 16x – 48 3x2 – 14x – 49 = 0 (3x + 7)(x – 7) = 0 Therefore, x = 7 as x = – 7—
3 is not acceptable.
2. 6x———–
( x – 1) = 5
Squaring both sides,
6x———–
( x – 1) = 25
6x = 25 x – 25 25 x = 6x + 25 Squaring both sides, 625x = (6x + 25)2
= 36x2 + 300x + 625 36x2 – 325x + 625 = 0 (4x – 25)(9x – 25) = 0
Therefore, x = 25—–4
or x = 25—–9
3. ( a + b c )2 = a + b2c + 2b ac Similarly,
( 18 + 3 )2 = 18 + 12(3) + 2(1) 18 × 3 = 21 + 216
Since 216 < 225 , ( 18 + 3 )2 < 21 + 225 ( 18 + 3 )2 < 36 18 + 3 < 6
4. 66 – 24 6 = p 2 + q 3 Squaring both sides, 66 – 24 6 = (p 2 + q 3 )2
= 2p2 + 3q2 + 2pq 6 Comparing the corresponding terms, 2pq = –24
q = –12—–p
......(1)
2p2 + 3q2 = 66
2p2 + 3144�——�p2 = 66
2p4 – 66p2 + 432 = 0 p4 – 33p2 + 216 = 0 (p2 – 9)(p2 – 24) = 0 p2 = 9 or p2 = 24 p = ±3 since p = ±2 6 is not an integer
Ace Ahead Mathematics S & T Volume 1 When p = ±3, q = �4 Hence, p = –3 and q = 4 since p = 3 and
q = –4 will give a negative value.
Therefore, 66 – 24 6 = –3 2 + 4 3
5. x1—3 – 3x
1– — 3 = 2
Let y = x1—3
y – 3y–1 = 2 y2 – 2y – 3 = 0 (y – 3)(y + 1) = 0 y = 3 or y = –1
x1—3 = 3 or x
1—3 = –1
x = 27 or x = –1
6. 27(32x) – 82(3x) + 3 = 0 Let y = 3x
27y2 – 82y + 3 = 0 (27y – 1)(y – 3) = 0 y = 1—–
27 or y = 3
3x = 1—–27
or 3x = 3
x = –3 or x = 1
7. 32x = 51 – x
2x log10 3 = (1 – x)log105 2x log10 3 + x log10 5 = log10 5 x(log10 3
2 + log10 5) = log10 5 x log10(9 × 5) = log10 5 x =
log10 5———–log10 45 = 0.6990———–
1.6532
= 0.4228 = 0.423 [3 signifi cant fi gures]
8. 42x – 8(22x) – 9 = 0 42x – 8(4x) – 9 = 0 Let y = 4x
y2 – 8y – 9 = 0 (y – 9)(y + 1) = 0 y = 9 or y = –1 4x = 9 since 4x = –1 is not possible x log10 4 = log10 9
x = log10 9———–log10 4
= 0.9542———–0.6021
= 1.58
[correct to 3 signifi cant fi gures]
9. (a) xyz = (loga b)(logb c)(logc a)
= (loga b)� loga c———loga b �� loga a———loga c � = loga a = 1 [shown]
(b) log10 (2n + 1) – log10 2n < log101.0025 log10 �2n + 1———
2n � < log10 1.0025 2n + 1———
2n < 1.0025
ACE STPM Math (Text Ans) 3rd.indd 1ACE STPM Math (Text Ans) 3rd.indd 1 3/27/2008 4:11:31 PM3/27/2008 4:11:31 PM
2
2n + 1 < (2n)1.0025 2n + 1 < 2.005n 2.005n – 2n > 1 0.005n > 1
n > 1——–0.005
n > 200 The least integral value of n is 201.
10. (a) log10 (a–1 + b–1) = log10 � 1—
a + 1—b �
= log10 � b + a——–ab �
= log10 (b + a) – log10(ab) = log10 (a + b) – log10 a
– log10 b [proven]
(b) logab x = logx x——––
logx ab
= 1——––———–
logx a + logx b 1 = ———————
loga a——–loga x
+ logb b——–logb x
1 = ———– 1—
y + 1—z
=
yz——–y + z
[proven]
11. log9 xy = 1—2
xy = 9
1—2
= 3
y = 3—x
Substitute y = 3—x into log3 x log3 y = –2,
log3 x log3 3—x = –2
log3 x[log3 3 – log3 x] = –2 log3 x – (log3 x)2 = –2 (log3 x)2 – log3 x – 2 = 0 Let p = log3 x, p2 – p – 2 = 0 (p – 2)(p + 1) = 0 p = 2 or p = –1 log3 x = 2 or log3 x = –1 x = 32 or x = 3–1
Therefore, x = 9, y = 1—3
or x = 1—3
, y = 9
12. 1 + 2i——–1 – i
= 1 + 2i——–1 – i
× 1 + i——–1 + i
= 1 + 2i + i + 2i2
——–————1 – i2
= –1 + 3i——–—2
Therefore, x = – 1—2
, y = 3—2
13. z1 = 3 – i z2 = 1 + i
1—z1
* + z1z2* =
1——–3 + i
+ (3 – i)(1 – i)
= 1——––
(3 + i)
(3 – i)——––(3 – i)
+ 3 – i – 3i + i2
= 3 – i——––9 + 1
+ 2 – 4i
= 3 – i——–10
+ 20 – 40i——–—
10
= 23—–10
– 41—–10
i
Therefore, a = 23—–10
and b = –41—–10
14. z1 = 1 + 5i z2 = 2 + i z1z2
* + z1
*z2
= (1 + 5i)(2 – i) + (1 – 5i)(2 + i) = 2 + 5 + 9i + 2 + 5 – 9i = 14 [shown]
15. (a) z = 2 – 2i z + z* = 2 – 2i + 2 + 2i = 4 z – z* = 2 – 2i – (2 + 2i) = –4i = –i(z + z*)[shown]
(b) 1—z + 1—
z* = 1——–
2 – 2i +
1——–2 + 2i
= � 1——–2 – 2i
× i—i � + � 1——–
2 + 2i × i—
i � =
i——–2i + 2
+ i——–
2i – 2
= i——–
2 + 2i –
i——–2 – 2i
= i� 1—z* – 1—
z � [shown] 16. (a) z1 = 4 – 3i z2 = 2 + i z3 = z1z2
z3 = (4 – 3i)(2 + i) = 8 + 3 – 2i = 11 – 2i |z3| = 112 + (–2)2
= 11.18
(b) arg z3 = tan–1 � –2—–11 �
= –0.18 radian (c)
17. 3 – ai
————1 – 3i
= 3 – ai
————1 – 3i
× 1 + 3i
————1 + 3i
= 3 + a 3 + (3 – a)i
—————————–1 + 3
= 3 (a + 1) + (3 – a)i
—————————–—–4
Im
ORe
z3 (11, –2)
ACE STPM Math (Text Ans) 3rd.indd 2ACE STPM Math (Text Ans) 3rd.indd 2 3/27/2008 4:11:33 PM3/27/2008 4:11:33 PM
3
If 3 – ai
————1 – 3i
is real number,
3 – a = 0 a = 3
Therefore, the real number = 3(3 + 1)
—————4
= 3
18. z1 = 1 – 3i z2 = 3 + i z1z2 = (1 – 3i)( 3 + i) = 3 + 3 + (1 – 3)i = 2 3 – 2i
r = |z1z2| = (2 3 )2 + (–2)2
= 12 + 4 = 4 θ = arg z1z2 = tan–1 � –2——–
2 3 � = –
π—6
Therefore, z1z2 = 4�cos �– π—6 � + i sin �– π—
6 ��19. (x + yi)2 = 4i x2 – y2 + 2xyi = 4i Comparing the real and imaginary parts, x2 – y2 = 0 ...... (1) 2xy = 4
y = 2—x
...... (2)
(2) into (1),
x2 – � 2—x �
2
= 0
x4 – 4 = 0 (x2 – 2)(x2 + 2) = 0
x = ± 2 since x = ± –2 does not have real values.
Therefore, x = ± 2 , y = ± 2
20. Let z = a + bi zz* – 5iz = (a + bi)(a – bi) – 5i(a + bi) = 10 – 20i a2 + b2 – 5ai + 5b = 10 – 20i (a2 + b2 + 5b) – 5ai = 10 – 20i
Comparing the real and imaginary parts, a2 + b2 + 5b = 10 ......(1) 5a = 20 a = 4
Substitute a = 4 into (1), 16 + b2 + 5b = 10 b2 + 5b + 6 = 0 (b + 3)(b + 2) = 0 b = –2 or b = –3
Therefore, z = 4 – 2i or z = 4 – 3i
21. (a) Let z = a – bi (a – bi)2 = 1 – 2 2i a2 – b2 – 2abi = 1 – 2 2i
Comparing the real and imaginary parts, a2 – b2 = 1 ......(1) 2ab = 2 2
b = 2
—–a
Substitute b = 2
—–a into (1),
a2 – 2—–a2 = 1
a4 – a2 – 2 = 0 (a2 – 2)(a2 + 1) = 0 a2 = 2 since a2 > 0 for real values of a. When a = ± 2 , b = ±1 Therefore, z1 = 2 – i and z2 = – 2 + i
(b)
(c) |z1| = ( 2 )2 + (–1)2
= 3
|z2| = (– 2 )2 + (1)2
= 3 arg z1 = –tan–1� 1—–
2 � = –0.615 radian arg z2 = π – tan–1� 1—–
2 � = 2.526 radians
22. z2 + 4z = 4 – 6i (z + 2)2 – 22 = 4 – 6i (z + 2)2 = 8 – 6i z + 2 = 8 – 6i Let (a + bi)2 = 8 – 6i a2 – b2 + 2abi = 8 – 6i Comparing the real and imaginary parts, a2 – b2 = 8 ...... (1) 2ab = –6
b = –3—–a
Substitute b = –3—–a into (1),
a2 – 9—–a2 = 8
a4 – 8a2 – 9 = 0 (a2 – 9)(a2 + 1) = 0 a2 = 9 since a2 > 0 for real values of a. When a = ±3, b = �1 Hence, z + 2 = 3 – i or z + 2 = –3 + i Therefore, z = 1 – i or z = –5 + i When z = 1 – i, |z| = 12 + (–1)2
= 2
Im
O Re
( 2, –1)
( –2, 1)
z1
z2
ACE STPM Math (Text Ans) 3rd.indd 3ACE STPM Math (Text Ans) 3rd.indd 3 3/27/2008 4:11:33 PM3/27/2008 4:11:33 PM
4
arg z = – tan–1 � 1—1 �
= – π—4
radian
When z = –5 + i, |z| = (–5)2 + 12
= 26
arg z = π – tan–1� 1—5 �
= 2.94 radians
23. n(A � B) = 36 n(A � B) = 4 n(B – A) = 3[n(A – B)] Let n(A – B) = x and n(B – A) = y = 3x x + 4 + y = 36 x + 4 + 3x = 36 4x = 32 x = 8 y = 24 n(A) = 8 + 4 = 12 n(B) = 24 + 4 = 28
24.
n(A) = 38 – (6 + 10 + 7) = 15
25. (a) If A � B, then A � B = B If B � C, then B � C = B Therefore, A � B = B � C [shown] (b) (A – B) � B = (A � B') � B = (A � B) � (B' � B) [distributive law] = (A � B) � ξ = A � B = A if B � A [shown]
26. (a) A � ( B � A') = (A � B) � (A � A') [distributive law] = (A � B) � ξ = A � B [proven]
(b) A – (A � B)' = A � (A � B) = (A � A) � (A � B) [distributive law] = A � (A � B) = A [proven]
27. (A � B) – (A � C) = (A � B) � (A � C)' = (A � B) � (A' � C') [de Morgan’s law] = (A � B � A') � (A � B � C') = [(A � A') � B] � [A � (B � C')]
[commutative law] = (φ � B) � [A � (B – C)] = φ � [A � (B – C)] = A � (B – C) [proven]
28. B – [(A � B)' – (A' � B)] = B – [(A � B)' � (A' � B)'] = B – [(A' � B') � (A � B')] [de Morgan’s law] = B – [(A' � A) � B'] [distributive law] = B – (φ � B') = B – B' = B � B = B
29. (a) A � (B' � C)' = A � (B � C') [de Morgan’s law] = (A � B) � C' [associative law] = (A � B) – C [proven] (b) A – (B � C) = A � (B � C)'
= A � (B' � C') [de Morgan’s law] = (A � B') � (A � C') [distributive law] = (A – B) � (A – C) [proven]
30. (A � B)' � [(A � C) – (B � C)] = (A � B)' � [(A � C) � (B � C)'] = (A' � B') � [(A � C) � (B' � C')] [de Morgan’s law] = (B' � A') � (B' � C') � (A � C)
[commutative law] = B' � (A' � C') � (A � C) [distributive law] = B' � (A � C)' � (A � C) [de Morgan’s law] = B' � φ = φ [proven]
A B
CD
A' � B � C'
3
6
7
n(C) = 10C' � D
A B
x y4
ACE STPM Math (Text Ans) 3rd.indd 4ACE STPM Math (Text Ans) 3rd.indd 4 3/27/2008 4:11:33 PM3/27/2008 4:11:33 PM