数学Paper 1讨论专区

Lov瑜瑜4ever

回复  Lov瑜瑜4ever

勉强来说还过得去, 看用什么角度去看...

有答案吗?
whyyie 发表于 18-7-2010 04:05 PM

就是没有答案
哈哈

芭樂

回复 1440# peaceboy


    surd of x-1... 哈哈
不好意思讓你看到Blurblur

peaceboy

回复  peaceboy


    surd of x-1... 哈哈
不好意思讓你看到Blurblur
芭樂 发表于 18-7-2010 08:39 PM

所以问题是

    if y = 1/[ [(x-1)^(1/2)] *5(sin x)], find dydx when x = pie /2?

首先let  [(x-1)^(1/2)] *5(sin x) = u
du/dx =  [(x-1)^(1/2)]5cosx + 5(sin x)[(1/2) [(x-1)^(-1/2)] ]
         =  [(x-1)(5 cos x ) + (5/2)(sin x)] / [(x-1)^(1/2)]

   then y = 1/u
dy/du = - 1/u^2
         = -1/ {[(x-1)^(1/2)] *5(sin x)}^2
          =-1/ 25(x-1)(sin x)^2

dy/dx = dy/du * du/dx
         =-1/ 25(x-1)(sin x)^2  * [(x-1)(5 cos x ) + (5/2)(sin x)] / [(x-1)^(1/2)]

    sub, x=pie /2
    懒惰算....大概是这样了

芭樂

回复 1443# peaceboy


Er...你曲解了
其實是1/[sqroots (x-1)] sin x
我朋友教我了
謝謝你的大恩大德

peaceboy

回复  peaceboy


Er...你曲解了
其實是1/[sqroots (x-1)] sin x
我朋友教我了
謝謝你的大恩大德
芭樂 发表于 18-7-2010 09:09 PM

    我多了一个5
不过做的方法也八九不离十...

harry_lim

walrein是天才来的
大家应该找他solve才对嘛

blazex

1)If p and q are positive numbers prove that
             (1-p)(1-q)>1-p-q
If p,q,and r are positive real numbers, with at least one of them less than unity,  
  prove that
            (1-p)(1-q)(1-r)>1-p-q-r

2)Express (√p + q√r)^2 in the form a+b√c .Without evaluating the square root  
   or using calculator, show that √10 + 2√2<6

whyyie

回复 1447# blazex

不明白这句"with at least one of them less than unity"

vick5821

1.Let m and n be positive integers, where m <=n. If the expansion of ( 1 + 2x)^n thr coefficient of x^(m+1) is 3/2 times the coefficient of x^m, show that and n satisfy 4n - 7m - 3 =0

谢谢

walrein_lim88

1.Let m and n be positive integers, where m
vick5821 发表于 26-7-2010 02:47 PM

vick5821

回复 1450# walrein_lim88

I dun understand the part how n! will finally got (n- m)??

peaceboy

回复  walrein_lim88

I dun understand the part how n! will finally got (n- m)??
vick5821 发表于 26-7-2010 05:15 PM

    他的方法shortcut很多东西

nC(m+1) = (n!)/[(n-m-1)!(m+1)!]
nCm = (n!)/[(n-m)!(m)!]

所以当
nC(m+1) = nCm
(n!)/[(n-m-1)!(m+1)!]= (n!)/[(n-m)!(m)!]
1/[(n-m-1)!(m+1)!] = 1/[(n-m)!(m)!]
（n-m)!/(n-m-1)! = (m+1)!/(m)!
（n-m)(n-m-1)!/(n-m-1)! = (m+1)(m!)/(m)!
所以  n-m=m+1

其他的部分就不用了吧

vick5821

回复 1452# peaceboy


    thanks you..can I have your msn address??

peaceboy

回复  peaceboy


    thanks you..can I have your msn address??
vick5821 发表于 26-7-2010 09:04 PM

    我很懒惰开msn的说 ... peace_loong@hotmail.com

vick5821

how he get 4 and 3 up there??

peaceboy

how he get 4 and 3 up there??
vick5821 发表于 26-7-2010 09:21 PM

扣除了 nC(m+1) 和 nCm

所以左边和右边剩下......
    2^(m+1) = (3/2) 2^m
       (2^m)(2^1) = (3/2) 2^m
                  2= 3/2
                  4 = 3

vick5821

回复 1456# peaceboy


    Ok..understand..thank you..I confused abt the coefficient of nth term and coefficient of x..apply same formula?

peaceboy

回复  peaceboy


    Ok..understand..thank you..I confused abt the coefficient of nth term and c ...
vick5821 发表于 26-7-2010 09:33 PM

    用binomial expension 的formula ...

http://www.youtube.com/watch?v=Cv4YhIMfbeM
http://www.youtube.com/watch?v=-fFWWt1m9k0&feature=channel

vick5821

I know..but how to differentitate which formula to use if they mention abt coefficient of x and another case is coefficient of term?

walrein_lim88

对不起。。。真的很忙。。。不过有人帮你解释就好了～～～