vol 1 chapter 1

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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)


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


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


vol 1 chapter 1

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