数学Paper 1讨论专区

白羊座aries

chapter 1,
帮我一个example 下.
1.using the laws of algebra for sets show that for any sets A and B
a) ( A n  B)' n A = B' n A
我做到:
A' u B' n A ..然后就停了,我觉得我错了,但是全部law中,只有De morgan's law 有compliment

2.If log4(base) N=p and log12(base) N = q ,show that
log3(base) 48=p+q/p-q

不会这题

3.Find the roots of the equation e^(2-2x) =2e^-x ,giving your answer in terms of logarithm.完全不会,答案对我来说也很特别.
Ans:2-In 2

4.Given a=14[1-(1/2)^x] and b=14 find the set of values of x such that the difference between a and b is less than 0.01,giving your answer correct to 2 d.p
      _________
5. /            __
   /  59-24/6       express as p square root 2 + q square root 3 where p and q are integer .由于太多square root,希望你们能明白.那题是 square root of 59-24square root of 6.

6.我不知道那里错.
express (1+i/1-2i) in form a+ib

(1+i/1-2i) x (1+2i/1+2i)

=(1+i)(1+2i)/(1-2i)(1+2i)
=(1-2i+i+2)/5
=(3-i)/5
=3/5 - 1/5 i
a=3/5 b = -1/5
但是答案和我相反 a= -1/5 b=3/5
然后arg怎样算啊?很乱下..为什么有些example是 pie - tan ^-1

7.z1 = 2+i ,z2= 2-i,z3= -1+2i .find modulus and argument

z3/z1

我是这样做:

(-1+2i)/(2+i) = (-1+2i )(2-i)/(2+i)(2-i)

=(-2+i+4i+2)/5
=5i/5
= i
不对咩?
答案是 1 , pie /2

早餐先,等下再战,晚上我要去旅行了啦....会的希望帮忙回答.谢谢

[ 本帖最后由 猪头鱼 于 2-5-2009 08:47 PM 编辑 ]

harry_lim

a) ( A n  B)' n A = B' n A

   (A n B)' n A
  =(A' U B') n A               [de morgan law]
  =(A' U A) n (B' n A)      [commutative law] <<<應該是這個law瓜？
  = Universal  n (B' n A)
  = B' n A                             (proved)

harry_lim

LZ太過干囧啦

SPM成績還沒出就。。。。。

harry_lim

express (1+i/1-2i) in form a+ib

= (1+i) / (1-2i) X (1+2i) / (1+2i)
= (1 + 2i + i + 2i^2) / (1 + 2i - 2i - 4i^2)
= (1 + 3i - 2) / (1+4)
= (3i - 1) / 5
= 3/5i - 1/5
a = -1/5 , b = 3/5

Ivanlsy

原帖由 白羊座aries 于 22-2-2009 11:13 AM 发表
If log4(base) N=p and log12(base) N = q ,show that
log3(base) 48=p+q/p-q

log_4 N = p
N = 4^p
(log_12 N) = q
N = 12^q

4^p = 12^q
plog_3 4 = qlog_3 12  
plog_3 4 = q[(log_3 4) + 1]
p(log_3 4) - q(log_3 4) = q
(log_3 4)(p - q) = q
log_3 4 = q/(p - q) --- 1

4^p = 12^q
(12/3)^p = 12^q
12^p / 12^q = 3^p
12^(p - q) = 3^p
(p - q)log_3 12 = p  
log_3 12 = p/(p - q) --- 2

1 + 2:
log_3 4 + log_3 12 = q/(p - q) + p/(p - q)
log_3 48 = (p + q)/(p - q)

Ivanlsy

原帖由 白羊座aries 于 22-2-2009 11:13 AM 发表
3.Find the roots of the equation e^(2-2x) =2e^-x ,giving your answer in terms of logarithm.完全不会,答案对我来说也很特别.
Ans:2-In 2

e^(2 - 2x) = 2(e^-x)
(e^2) / (e^2x) = 2 / (e^x)
e^2 = 2(e^x)
e^x = (e^2) / 2
x ln e = ln [(e^2) / 2]
x = 2ln e - ln 2
x = 2 - ln2

* ln = log_e

白羊座aries

原帖由 Ivanlsy 于 22-2-2009 12:37 PM 发表


log_4 N = p
N = 4^p
(log_12 N) = q
N = 12^q

4^p = 12^q
plog_3 4 = qlog_3 12  
plog_3 4 = q[(log_3 4) + 1]
p(log_3 4) - q(log_3 4) = q
(log_3 4)(p - q) = q
log_3 4 = q/(p - q) --- 1

4 ...

喷血...不明白....为什么有+1的?
我中五学的都没有那么复杂

白羊座aries

原帖由 Ivanlsy 于 22-2-2009 12:46 PM 发表

e^(2 - 2x) = 2(e^-x)
(e^2) / (e^2x) = 2 / (e^x)
e^2 = 2(e^x)
e^x = (e^2) / 2
x ln e = ln [(e^2) / 2]
x = 2ln e - ln 2
x = 2 - ln2

* ln = log_e

不明白了

Ivanlsy

原帖由 白羊座aries 于 22-2-2009 11:13 AM 发表
4.Given a=14[1-(1/2)^x] and b=14 find the set of values of x such that the difference between a and b is less than 0.01,giving your answer correct to 2 d.p

a = 14[1 - (1/2)^x]
   = 14 - 14(1/2)^x
b = 14

b - a < 0.01
14(1/2)^x < 0.01
(1/2)^x < 1/1400
2^-x < 1/1400
-x < log_2 (1/1400)
-x < -10.45
x > 10.45

Ivanlsy

原帖由 白羊座aries 于 22-2-2009 12:48 PM 发表


喷血...不明白....为什么有+1的?
我中五学的都没有那么复杂

(log_3 12) = [log_3 (4X3)]
                   = (log _3 4) + (log_3 3)
                   = log_3 4 + 1

白羊座aries

原帖由 Ivanlsy 于 22-2-2009 12:52 PM 发表


a = 14[1 - (1/2)^x]
   = 14 - 14(1/2)^x
b = 14

b - a < 0.01
14(1/2)^x < 0.01
(1/2)^x < 1/1400
2^-x < 1/1400
-x < log_2 (1/1400)
-x < -10.45
x > 10.45

-ve sign呢?为什么是positive?

Ivanlsy

原帖由 白羊座aries 于 22-2-2009 12:52 PM 发表


不明白了

(e^2) / (e^2x) = 2 / (e^x)
e^2 = 2(e^2x) / (e^x)
e^2 = 2[e^(2x - x)]
e^2 = 2(e^x)

白羊座aries

原帖由 Ivanlsy 于 22-2-2009 12:55 PM 发表

(log_3 12) = [log_3 (4X3)]
                   = (log _3 4) + (log_3 3)
                   = log_3 4 + 1

ok,我明白了

harry_lim

白羊。。。慢慢來啦

Ivanlsy

原帖由 白羊座aries 于 22-2-2009 12:56 PM 发表


-ve sign呢?为什么是positive?

b - a = 14 - [14 - 14(1/2)^x]
         = 14 - 14 + 14(1/2)^x
         = 14(1/2)^x

白羊座aries

原帖由 Ivanlsy 于 22-2-2009 12:58 PM 发表

b - a = 14 - [14 - 14(1/2)^x]
         = 14 - 14 + 14(1/2)^x
         = 14(1/2)^x

哦,明了明了.
顺便:
prove
1)A' n ( A U B' ) n C'=A' n B' n C'
2)( A-B ) U ( B-A )= ( A U B ) - ( A n B)
solve
3)(x+iy)^2 = i,find all the values of x and y .
BY completing the square or otherwise ,solve
x^2 +4x = -9 +12i

Ivanlsy

原帖由 白羊座aries 于 22-2-2009 11:13 AM 发表
5. √(59 - 24√6)express as p√2 + q√3 where p and q are integer.

let √(59 - 24√6) = p√2 + q√3
          59 - 24√6 = (p√2 + q√3)^2
                            = 2p^2 + 2pq√6 + 3q^2

Compare LHS and RHS of the equation:
2p^2 + 3q^2 = 59
2pq = -24
pq = -12
q = -12/p

2p^2 + 3(-12/p)^2 = 59
2p^2 + 3(144/p^2) = 59
2p^4 + 3(144) = 59p^2
2p^4 - 59p^2 + 432 = 0
Solving, get
p^2 =  16  or  p^2 = 13.5 (No statisfy)
p = ± 4

When p = 4, q = -12/4 = -3
When p =-4, q = -12/-4 = 3
Checking, we found that p = -4, q = 3 no statisfy with the equation,
Hence  √(59 - 24√6) = 4√2 - 3√3

[ 本帖最后由 Ivanlsy 于 22-2-2009 01:23 PM 编辑 ]

白羊座aries

原帖由 Ivanlsy 于 22-2-2009 01:19 PM 发表

let √(59 - 24√6) = p√2 + q√3
          59 - 24√6 = (p√2 + q√3)^2
                            = 2p^2 + 2pq√6 + 3q^2

Compare LHS and RHS of the equation:
2p^2 + 3q^2 = 59
2pq = -24
pq  ...

那里来的?

Ivanlsy

原帖由 白羊座aries 于 22-2-2009 11:13 AM 发表
7.z1 = 2+i ,z2= 2-i,z3= -1+2i .find modulus and argument

z3/z1

z_3 / z_1 = (-1 + 2i)/(2 + i)
                = (-1 + 2i)/(2 + i) X (2 - i)/(2 - i)
                = (-1 + 2i)(2 - i) / (4 + 1)
                = (-2 + i + 4i + 2) / 5
                = i
| z_3 / z_1 | = √(0^2 + 1^2)
                    = √1
                    = 1
arg ( z_3 / z_1 ) = tan^-1 (1/0)
                            = π/2

Ivanlsy

原帖由 白羊座aries 于 22-2-2009 01:26 PM 发表


那里来的?

(p√2 + q√3)^2 =  (p√2 + q√3)(p√2 + q√3)
                         = 2p^2 + pq√6 + pq√6 + 3q^2
                         = 2p^2 + 2pq√6 + 3q^2