佳礼资讯网

 找回密码
 注册

ADVERTISEMENT

楼主: 白羊座aries

数学Paper 1讨论专区

   关闭 [复制链接]
发表于 24-7-2009 10:42 PM | 显示全部楼层

回复 360# 末日审判 的帖子

有答案吗?怕做错~
回复

使用道具 举报


ADVERTISEMENT

 楼主| 发表于 24-7-2009 10:56 PM | 显示全部楼层
原帖由 末日审判 于 24-7-2009 09:15 PM 发表
帮帮我下,
1.The sum of the first four terms of a G.P is five times the sum of the sum of the first two terms.Find the value of the common ratio,r,given that r>1.

2.The 1st term of a G.P. exceeds  ...


我想看高手做第一题

2. a-ar = 1
a(1-r)=1
a=1/1-r  ---1
S3=13/5

a(1-r^3)/1-r =13/5 ----2

substitute 1 into 2
(1-r)(1-r^3)/1-r =13/5

然后你做玩后,会有一个cubic equation出来,然后你factorise 后就是答案了.

r=2/5 ;a=5/3  or
r= -4 ; a=1/5
回复

使用道具 举报

发表于 24-7-2009 11:00 PM | 显示全部楼层
question 1 use a(r^n -1)/r-1.
then u will solve until the whole equation turn into quadratic eqn.
回复

使用道具 举报

发表于 24-7-2009 11:02 PM | 显示全部楼层

回复 360# 末日审判 的帖子

a)let GP =a,ar,ar^2,and sum of the first n term of series=Sn
given that S4=5S2
a(1-r^4)/1-r=5a(1-r^2)/1-r
          1-r^4=5(1-r^2)
(1+r^2)(1-r^2)=5(1-r^2)
                1+r^2=5
                    r^2=4
                      r=2
回复

使用道具 举报

发表于 24-7-2009 11:07 PM | 显示全部楼层
Sum of 1st 4T = a(r^4 - 1)/(r - 1)       ...........................A
Sum of 1st 2T = a(r^2 - 1)/(r - 1)       ...........................B
A = 5B
(r^4 - 1) = 5(r^2 - 1)
r^4 - 5r^2 + 4 = 0
let r^2 =x
x^2 - 5x + 4= 0
x = 1, 4
then ,  r =  1, -1, 2, -2
since r >1, then r = 2
回复

使用道具 举报

发表于 24-7-2009 11:08 PM | 显示全部楼层
opssss
回复

使用道具 举报

Follow Us
发表于 24-7-2009 11:09 PM | 显示全部楼层

回复 365# 四月一日的小皮 的帖子

same answer
回复

使用道具 举报

发表于 25-7-2009 09:59 AM | 显示全部楼层
原帖由 白羊座aries 于 24-7-2009 10:56 PM 发表


我想看高手做第一题

2. a-ar = 1
a(1-r)=1
a=1/1-r  ---1
S3=13/5

a(1-r^3)/1-r =13/5 ----2

substitute 1 into 2
(1-r)(1-r^3)/1-r =13/5

然后你做玩后,会有一个cubic equation出来,然后你 ...
原帖由 ynnee 于 24-7-2009 11:02 PM 发表
a)let GP =a,ar,ar^2,and sum of the first n term of series=Sn
given that S4=5S2
a(1-r^4)/1-r=5a(1-r^2)/1-r
          1-r^4=5(1-r^2)
(1+r^2)(1-r^2)=5(1-r^2)
                1+r^2=5
           ...
原帖由 四月一日的小皮 于 24-7-2009 11:07 PM 发表
Sum of 1st 4T = a(r^4 - 1)/(r - 1)       ...........................A
Sum of 1st 2T = a(r^2 - 1)/(r - 1)       ...........................B
A = 5B
(r^4 - 1) = 5(r^2 - 1)
r^4 - 5r^2 + 4 = 0
let r^ ...

你们答案都对了,很谢谢你们!
你们用来解决问题的方式都不同,我也学到了!
谢谢!
回复

使用道具 举报


ADVERTISEMENT

发表于 26-7-2009 02:55 PM | 显示全部楼层
我又有新問題了,希望大家能幫忙!
1.The first three of four integers are in A.P.and the last three are in a G.P.Find these four numbers given that the sum of the first and the last integers is 37 and the sum of the two integers in the middle is 36.(answer:12,16,20,25)
2.The 2nd,6th and 8th terms of an A.P.are three successive terms of a G.P.Find the common ratio of the G.P.and obtain an expression for the nth term of the G.P.(answer:r=1/2;16/9  a(1/2)^n
3.Prove that 3^x,3^y,and3^z are successive terms of a G.P.,if x,y,and z are successive terms of an A.P.
回复

使用道具 举报

发表于 26-7-2009 09:56 PM | 显示全部楼层
1. let ap is (a),(a+b),(a+2b) for 1st 3 term.
then for 4th integer is (a+2b)^2 / (a+b)  ------>common ratio come from term 2 and term 3 , since last 3 term is gp
term 2 + term 3 = a + b + a + 2b = 36
                                    2a  + 3d  = 36
                                               d  =  12  - 3a/2   ..............................1
term 1 + term 4 = 37
.........
(i simplify the eqn to become this)
a(a + d) + (a + d)^2 = 37 (a + d).....................................................2
subtitute eqn 1 into 2 :
.....(again, i simplify it since it too complex to write out )
.......
4a^2/9  -  49a / 3  + 132  = 0
(then do like quadratic eqn)
a = 12, 99/4
when a = 12
         d =  12 - 2(12)/3  = 4
thus, term 1 = a = 12
        term 2 = a + d = 12 + 4 = 16
        term 3 = a + 2d = 20
       term 4 = (20)^2  / 16 = 25         (look to the common ratio eqn )

i quite not sure whether my working is follow skema  since my answer gt 2 and the value 99/4 is oso can use into the series. if any shorter working pls show me
回复

使用道具 举报

发表于 26-7-2009 10:03 PM | 显示全部楼层
from the ap, the 3 following gp  is shown :
a+d , a+ 5d ,a +7d
since common ratio is same,then
(a+5d)/(a+d) = (a+ 7d)/(a+5d)
(a+5d)^2 = (a+7d)(a +d)
a^2 + 10ad + 25d^2 = a^2 + 8ad + 7d^2
                      2ad   =  - 18 d^2
                           a  =  -9 d
thus,   r = ( -9d + 5d) / (-9d + d)
             =  -4d / -8d    = 1/2
the gp can form the following series using d= -a/9
(8/9)a,(4/9)a,(2/9)a.....
1st term = 8a/9
eqn ----->  term n = (8/9)a (1/2)^(n-1)
                            = (8/9)(1/2)a(1/2)^n
                            =  16/9  a(1/2)^n

回复

使用道具 举报

发表于 26-7-2009 10:06 PM | 显示全部楼层
x,y,z is ap thus
y - x = z - y
let m = y -x = z - y
then, 3^m = 3^(y - x) = 3 ^(z -y)
                     3^y / 3^x  =  3^z / 3^y.............> common ratio
thus 3^x , 3^y , 3^z is a gp series.
(correct me if i was wrong.... )
回复

使用道具 举报

发表于 26-7-2009 11:17 PM | 显示全部楼层
  2a  + 3d  = 36
            d  =  12  - 3a/2   ..............................1

d不是等于d=12-2a/3吗?

term 1 + term 4 = 37
.........
(i simplify the eqn to become this)
a(a + d) + (a + d)^2 = 37 (a + d).....................................................2
subtitute eqn 1 into 2 :
.....(again, i simplify it since it too complex to write out )
.......

这边不明白,可否详细点?
回复

使用道具 举报

发表于 27-7-2009 05:42 PM | 显示全部楼层
原帖由 末日审判 于 26-7-2009 11:17 PM 发表

d不是等于吗?


这边不明白,可否详细点?



ummm.....yes, is  d=12-2a/3...mistake.
term 1 = a
term 4 = (a + 2d)^2 / (a + d)
then...
a + (a + 2d)^2 / (a + d) = 37
(multiply both side with (a+d):
a(a+d) + (a+ 2d)^2 = 37(a+d)
回复

使用道具 举报

发表于 6-8-2009 06:41 PM | 显示全部楼层
∫ ln x dx 該怎麽solve?
:@

我算到要吐血了....
回复

使用道具 举报

发表于 6-8-2009 08:54 PM | 显示全部楼层
原帖由 Dicardo 于 6-8-2009 06:41 PM 发表
∫ ln x dx 該怎麽solve?
:@

我算到要吐血了....



你可以用Integration by part.

我额外补充一下。当你们用integration by part时,
你们得知道一些 rule,就是选出 first function当成你们的fixed function.
例如:
∫   f(x). g(x)  dx  = f(x) ∫ g(x) dx - ∫  [ ∫ g(x) dx ] . f '(x)  dx


In words, this formula states that

   The integral of the product of two functions

= first function x integral of second - integral of  ( integral of second  x diff. coefficient of first )


所以你们先在两个functions中锁定其中一个function为你的first function(我习惯用Fixed这个字眼).

其实,这里有个我个人的口诀。

By using integration by part,

Fixed . Integrate  -  
Integrate . differentiate Fixed   dx



接下来,我教教你们如何 辨认 哪个 functions 为 你们的 first function.

这里就是要obey the rules 来找出
first function。

The first function is chosen as the function which comes first in the word ILATE, where

I    -    stands for Inverse function              (Exp:  f(x) = sin^(-1) x; cos^(-1) )
L  -    stands for Logarithmic function      
(Exp:  f(x) = log x;    ln x )
A  -    stands for Algebraic function          (Exp:  f(x) =  2x+3 )
T  -    stands for  Trigonometric function
(Exp:  f(x) =  sin x; cos x )
E  -    stands for Exponential function      (Exp:  f(x) =  e^(x) )


通常STPM的level, 只有以下的rules:

lnx
x
e^x


你们根据以上的rules就可以找出fixed 就是 (first function)。

回你的问题:

∫ ln x dx

这个问题有两个functions就是

     ∫ ln x dx
=   ∫ 1  .  ln x dx

设 f(x) = ln x 和 g(x) = 1

你现在可以很清楚地 辨认 哪一个是 fixed 吗?
就是用刚才我所说的rules来找出fixed.

其实 ln x 是 属于 Logarithmic function,
而 1  是 属于 Algebraic function

根据rules:
lnx
x
e^x


你看到lnx 排在 x 的上面,所以 lnx 就是 设为 Fixed.

然后根据这个

∫   f(x). g(x)  dx  = f(x) ∫ g(x) dx - ∫  [ ∫ g(x) dx ] . f '(x)  dx



∫   f(x). g(x)  dx = Fixed . Integrate  -  Integrate . differentiate Fixed   dx

Solution:

f(x) = ln x 为 Fixed

g(x) = 1    为 second

     ∫ ln x dx
=   ∫ 1  .  ln x dx
=   ln x . x -     x . 1/ x  dx         {因为 Fixed . Integrate  -  Integrate . differentiate Fixed dx }
=   x.lnx    -    1       dx   
=   x.lnx    -   x  +   c
=   x ln (x/e)      +   c
回复

使用道具 举报


ADVERTISEMENT

 楼主| 发表于 10-8-2009 10:59 PM | 显示全部楼层
by substituting y= x + 1/x

f(x)= x^3  -4x -6 -4/x +1/(x^3)

y^2-2=x^2+1/x^2

然后要怎样
回复

使用道具 举报

发表于 10-8-2009 11:43 PM | 显示全部楼层
原帖由 白羊座aries 于 10-8-2009 10:59 PM 发表
by substituting y= x + 1/x

f(x)= x^3  -4x -6 -4/x +1/(x^3)

y^2-2=x^2+1/x^2

然后要怎样


f(x) = x^3 + 1/(x^3) - 4x - 4/x - 6
       = x^2 + 1/x^2 - 4 - 4/x^2 - 6/x (divide x)
      =(y^2 - 2) - .............(我忘记了)
回复

使用道具 举报

发表于 11-8-2009 12:59 AM | 显示全部楼层
原帖由 白羊座aries 于 10-8-2009 10:59 PM 发表
by substituting y= x + 1/x

f(x)= x^3  -4x -6 -4/x +1/(x^3)

y^2-2=x^2+1/x^2

然后要怎样


f(x) = x^3 - 4x - 6 - 4/x + 1/(x^3)

f(x) = x^3  + 1/(x^3)  - 4x - 4/x -6


since y= x + 1/x
=>   y^3 = (x + 1/x)^3
=>   y^3 = [ x^3 + 1/(x^3) ] + 3(x + 1/x)
=>   y^3 = [ x^3 + 1/(x^3) ] + 3y
=>  y^3 -3y = [ x^3 + 1(/x^3) ]

Therefore,
f(x) = x^3  + 1/(x^3)       - 4x - 4/x    -6
f(x) =[ x^3  + 1/(x^3) ] - 4(x + 1/x) -6

f(y) = [y^3 -3y] - 4y -6
f(y) =  y^3 -7y - 6
回复

使用道具 举报

发表于 11-8-2009 01:03 AM | 显示全部楼层
原帖由 harry_lim 于 10-8-2009 11:43 PM 发表


f(x) = x^3 + 1/(x^3) - 4x - 4/x - 6
       = x^2 + 1/x^2 - 4 - 4/x^2 - 6/x (divide x)
      =(y^2 - 2) - .............(我忘记了)


你为什么无端端 (divide x)
解释一下。
回复

使用道具 举报

您需要登录后才可以回帖 登录 | 注册

本版积分规则

 

所属分类: 欢乐校园


ADVERTISEMENT



ADVERTISEMENT



ADVERTISEMENT

ADVERTISEMENT


版权所有 © 1996-2023 Cari Internet Sdn Bhd (483575-W)|IPSERVERONE 提供云主机|广告刊登|关于我们|私隐权|免控|投诉|联络|脸书|佳礼资讯网

GMT+8, 14-1-2025 08:23 PM , Processed in 0.143837 second(s), 20 queries , Gzip On.

Powered by Discuz! X3.4

Copyright © 2001-2021, Tencent Cloud.

快速回复 返回顶部 返回列表