数学Paper 1讨论专区

peaceboy

given that the equation x^2 -(a+2)x + (a+5)=0 has positive roots u and v where both are less than 5. ...
數學神童 发表于 5-10-2010 02:48 PM

    0<v<5 , 0<u<5
0<u+v<10
0<a+2<10
-2<a<8

0<uv<25
0<a+5<25
-5<a<20

{a : -2<a<8}
不懂对没有

數學神童

but when i substitute a=0 inside it does not satisfy the condition given so i don think ur answer is correct

peaceboy

but when i substitute a=0 inside it does not satisfy the condition given so i don think ur answer is ...
數學神童 发表于 5-10-2010 05:20 PM

    我就知道不可能这么容易 - -

b^2-4ac >0
[-(a+2)]^2 -4(1)(a+5)>0
a^2+4a+4-4a-20>0
a^2-4^2>0
(a+4)(a-4)>0
a<-4  , a>4


    0<v<5 , 0<u<5
0<u+v<10
0<a+2<10
-2<a<8

0<uv<25
0<a+5<25
-5<a<20


{a : 4<a<8}

其实有答案么？做到很不确定，呵呵

kelfaru

我只知道这个叫fallacy
数学是讲逻辑的，1-1+1+.....答案不是0就是1，肯定不会是0.5
就 ...
peaceboy 发表于 5-10-2010 04:56 PM

话也许不可这么说~
看看wiki...
http://en.wikipedia.org/wiki/Grandi%27s_series
从wikil里边分成telescoping series和divergent geometry series,
到最后好像都没一个定论~

说是0或1没有意义,因为infinity没有一个定论~

kelfaru

这边有哪位大大可以上载今年STPM trial paper 1,各州的,小弟的TRIAL　PAPER 是在72页ALLMATHS大大所POST的,
小弟在此谢过先～

peaceboy

回复 1904# kelfaru


On the other hand, its Cesàro sum is 1/2.

Cesàro summation is an averaging method, in that it relies on the arithmetic mean of the sequence of partial sums.

很好，你知道没有意思就好，你不能说他是0，不能说他是1，但也不是因此就可以说他是1/2

同理1-2+4-8+....to infinity 也是一样没有意思的

kelfaru

回复 1906# peaceboy


小弟说的是telescoping series和divergent geometry series，

• The series 1 &#8722; 1 + 1 &#8722; 1 + … has no sum.
  
  
  
  
• ...but its sum should be 1/2.

没有一个定论～

peaceboy

很好，所以你不能说他是0，不能说他是1，但也不是因此就可以说他是1/2．．．
有意见么？

數學神童

find the restriction of x if triangle ABC exists in such a way that AB=3/(x+3) cm , BC=2/(x-1) cm and the longest side AC=4/(x-4) cm

Allmaths

find the restriction of x if triangle ABC exists in such a way that AB=3/(x+3) cm , BC=2/(x-1) cm an ...
數學神童 发表于 6-10-2010 02:46 PM

By triangle inequality,
AB+BC>AC
3/(x+3)+2/(x-1)>4/(x-4)
(5x+3)/(x+3)(x-1)-4/(x-4)>0
[x(x-25)]/[(x+3)(x-1)(x-4)]>0

{x: -3<x<0 or 1<x<4 or x>25}
But x>4,

∴{x: x>25}


不知道对不对...


上次没注意到x一定要大过4...

peaceboy

回复 1909# 數學神童


    好像要用到form 5的trianlge 的东西，angle和side那个chapeter = =

大概就是 最短哪两条线的angle between 0 到 pi瓜
不是很记得formula了......得空再翻翻书看 {:2_72:}

这不是STPM的问题吧 {:2_81:}

Jacss

大家酱得空,给大家一题,从别边拿来的

1 - 1 + 1 - 1 + 1 - 1 + ...... (到 infinity) = ?
有两个答案 ...
kelfaru 发表于 4-10-2010 11:27 PM

This is one type of alternating series, and certainly this is beyond the syllabus of STPM paper.

The answer 1/2 is obtained when using that averaging method, which involved partial summation and some assumptions.


Normally mathematician only interested in whether the series above is converging or diverging, and not the value.

weitao

这题怎样做？
Show that a^2+b62>or=2ab
Ifx+y+z+c,show that x^2+y^2+z^2>or=(c^2)/3

peaceboy

这题怎样做？
Show that a^2+b62>or=2ab
Ifx+y+z+c,show that x^2+y^2+z^2>or=(c^2)/3
weitao 发表于 7-10-2010 03:54 PM

    (a-b)^2 >=0
a^2+b^2 - 2ab>=0
a^2+b^2>= 2ab

x+y+z=c,
c^2 =(x+y+z)^2
      =(x+y+z)(x+y+z)
      =x^2+xy+xz +xy+y^2+yz+xz+yz+z^2
      =x^2+y^2+z^2 +2xy+2xz+2yz


x^2+y^2 + y^2 +z^2 + z^2+x^2 >= 2xy+2yz+2xz
2x^2 + 2y^2 +2z^2 >= 2xy+2yz+2xz
3x^2 + 3y^2 + 3z^2 >= x^2+y^2+z^2 + 2xy+2yz+2xz
3(x^2+y^2+z^2)>= c^2
x^2+y^2+z^2 >= c^2 /3

數學神童

given that a quadratic equation is x^2 + (k-3)x + k =0 .Find the set of values of k if the equation has
i)2 real roots
ii)2 roots with same sign

peaceboy

given that a quadratic equation is x^2 + (k-3)x + k =0 .Find the set of values of k if the equation  ...
數學神童 发表于 7-10-2010 04:39 PM

    i) b^2-4ac >=0 ,
(k-3)^2 -4k>=0
k^2 -6k +9 -4k>=0
k^2 -10k +9>=0
(k-1)(k-9)>=0
k<=1 ,k>=9

ii)  for the root to be the same sign , k>=0
0=< k=<1  , k>=9

peaceboy

This is one type of alternating series, and certainly this is beyond the syllabus of STPM paper. ...
Jacss 发表于 7-10-2010 03:24 PM

    你是大学的？

Jacss

回复 1917# peaceboy


   haha.. 被拆穿了，不过我曾经也是form6的学生..

peaceboy

回复  peaceboy


   haha.. 被拆穿了，不过我曾经也是form6的学生..
Jacss 发表于 7-10-2010 07:22 PM

    原来是学长
很好奇的问下，你主修数学？

Jacss

回复 1919# peaceboy


   哈哈..可以算是吧..