数学Paper 1讨论专区

Allmaths

有一题我也跟着他方法但做不到
  3x^2+px+1=0and2x^2+qx+1=0have a common root.
  show that  ...
hongji 发表于 6-12-2010 10:44 PM

   
3x^2+px+1=0   --->eq 1
2x^2+qx+1=0   --->eq 2

2(eq 1)-3(eq 2),

6x^2+2px+2-(6x^2+3qx+3)=0
(2p-3q)x-1=0
x=1/(2p-3q)

Sub x=1/(2p-3q) into eq 1,
3[1/(2p-3q)]^2+p[1/(2p-3q)]+1=0
3+p(2p-3q)+(2p-3q)^2=0
2p^2+3q^2-5pq+1=0   (shown)

hongji

3x^2+px+1=0   --->eq 1
2x^2+qx+1=0   --->eq 2

2(eq 1)-3(eq 2),

6x^2+2px+2-(6x^2+3 ...
Allmaths 发表于 6-12-2010 10:59 PM

不好意思问这么多问题
我还有很多问题呢〉《
complex number z1 and z2 are given by z1=5+i
                                                        z2=2-3i
determine the value of the real constants @ and β such that

@+iβ +3z1 / @ -iβ +3z2=2i

junchung2003

回复 2302# hongji


    问题是什么？看不清

hongji

回复  hongji


    问题是什么？看不清
junchung2003 发表于 7-12-2010 03:50 PM

    The complex number Z1 and Z2 are given by Z1= 5+i and Z2=2-3i.
    Determine he values of the real constants @ and β such that

  [@+iβ+3zi] / [@-iβ+3z2]  = 2i

junchung2003

回复 2304# hongji


   
LHS
= [@+iβ+3zi] / [@-iβ+3z2]
= [@+iβ+15+3i] / [@-iβ+6-9i]
= [(@+15)+(3+β)i]/[(@+6)-(β+9)i]

[(@+15)+(3+β)i]/[(@+6)-(β+9)i] = 2i
[(@+15)+(3+β)i] = [(@+6)-(β+9)i] * 2i
[(@+15)+(3+β)i] = 2i(@+6) + 2(β+9)
(@+15)+(3+β)i = 2(β+9) + (2@+12)i

Compare both side,

@+15 = 2β+18
@ = 2β+3

3+β = 2@+12
3+β = 2(2β+3)+12
3+β = 4β+6+12
3β = -15
β = -5

Subs. β=-5 into @ = 2β+3

@ = 2(-5)+3
@ = -7

@ = -7 , β = -5

hongji

回复  hongji


   
LHS
= [@+iβ+3zi] / [@-iβ+3z2]
= [@+iβ+15+3i] / [@-iβ+6-9i]
= [(@+15) ...
junchung2003 发表于 7-12-2010 08:29 PM

    为什么x3先才拿到答案啊？

junchung2003

回复 2306# hongji


    那里x3？

hongji

回复  hongji


    那里x3？
junchung2003 发表于 7-12-2010 09:58 PM

[@+iβ+5+i] / [@-iβ+2-3i]
乘3 后得 [@+iβ+15+3i] / [@-iβ+6-9i]

junchung2003

回复 2308# hongji


我跳了

= [@+iβ+3zi] / [@-iβ+3z2]
= [@+iβ+3(5+i)] / [@-iβ+3(2-3i)]   --> 把Z1和Z2Substitute进去罢了

= [@+iβ+15+3i] / [@-iβ+6-9i]
= [(@+15)+(3+β)i]/[(@+6)-(β+9)i]

hongji

回复  hongji


我跳了

= [@+iβ+3zi] / [@-iβ+3z2]
= [@+iβ+3(5+i)] / [@-iβ+3(2-3i)]   --> 把 ...
junchung2003 发表于 7-12-2010 11:21 PM

    对呢。。我都没看到
不好意思

whyyie

Using a suitable substitution, integrate 3x / (4 + x^4) with respect to x.

junchung2003

ʃ 3x/(4+x^4) dx

let
x^2 = 2 tan ϴ
(x^2)/2 = tan ϴ
ϴ = tan^-1 (x^2/2) --- (1)

2x(dx/dϴ ) = 2 sec^2 ϴ

dx = (sec^2 ϴ/x) * dϴ

ʃ 3x/(4+x^4) dx
= ʃ [3x/(4+(2 tan ϴ )^2)] * (sec^2 ϴ/x) * dϴ
= ʃ [(3 sec^2 ϴ )/ 4(1+tan^2 ϴ )] * dϴ
= ʃ 3/4 * dϴ --> 1+tan^2 ϴ = sec^2 ϴ
= 3ϴ/4 +C
= (3/4) * tan^-1 (x^2/2) +C --> from (1)

whyyie

回复 2312# junchung2003

答案给 (3/2) tan-1 (x^2)

junchung2003

有没有给你step？
也许我还不够强，只是你那个答案不怎么可能

Log

回复 2314# junchung2003
放心。。。对了

junchung2003

回复 2315# Log


    呵呵
谢谢 我还担心自己那里careless呢

Log

如果是  integrate 3x / (1 + x^4) with respect to x.

那么答案就是 (3/2) tan^-1 (x^2) +c

hongji

find the sum of the series
[1/2.3.4+2/3.4.5+3/4.5.6+...+n/(n+1)(n+2)(n+3)]

Ur=1+2+3/([2+(n-1)1][3+(n-1)1][4+(n-1)1])  《〈他的分子对吗？

junchung2003

回复 2318# hongji


    找partial fraction先
然后才Sum to n term

whyyie

The equation of a curve is y√x - x√y + x^2 = 3.
Find the y-coordinate of the point on the curve where x=1 and the gradient of the curve at that point.