数学Paper 1讨论专区

whyyie

simplify  [ (1 + 2√3)^5 ]  -[ (1 - 2√3)^5 ]


这个哦，是不是这样做的。。。
[(5C0(1)^0(2√3)^5  + 5C1(1)^1(2√3)^4  + 5C2(1)^2(2√3)^3  + 5C3(1)^3(2√3)^2  + 5C4(1)^4(2√3)^1  + 5C5(1)^5(2√3)^0 ] - [(5C0(1)^0(-2√3)^5  + 5C1(1)^1(-2√3)^4  + 5C2(1)^2(-2√3)^3  + 5C3(1)^3(-2√3)^2  + 5C4(1)^4(-2√3)^1  + 5C5(1)^5(-2√3)^0 ]

酱的话等于零吗？
hongji 发表于 20-9-2011 08:59 PM

= 2 * 5C0(1)^0(2√3)^5 + 2 * 5C2(1)^2(2√3)^3 + 2 *  5C4(1)^4(2√3)^1
= 2 [ 5C0(1)^0(2√3)^5 +  5C2(1)^2(2√3)^3 +  5C4(1)^4(2√3)^1 ]
= 2 [ (2√3)^5 + 10(2√3)^3 + 5(2√3) ]
= 4√3 [(2√3)^4 + 10(2√3)^2 + 5]
= 4√3 (144 + 120 + 5)
= 1076√3

hongji

= 2 * 5C0(1)^0(2√3)^5 + 2 * 5C2(1)^2(2√3)^3 + 2 *  5C4(1)^4(2√3)^1
= 2 [ 5C0(1)^0(2√3)^5 +  ...
whyyie 发表于 20-9-2011 10:56 PM

为什么不用减的？可以告诉我吗？

hongji

equation ax^2 + bx + c =0 with a,b and c as non-zero constants, has roots α and β.

1)if α=入β ,with 入=/= 0 and 入=/=1 ,and equation lx^2 +mx + n =0 with l,m and n as non-zero constants , has roots 入(入+1)^2 and (入+1)^2 .show that [(m^2)/ (ln)] = [(b^2 )/ (ac)]



2)if α>β and equation x^2 + px +q =0 with p and q as non-zero constants has roots α^2 + β^2  and   α^2 - β^2 ,express p and q in terms a, b and c.

hongji

the first and the second term of an arithmetic series equal respectively the first and the second term of a geometric series. The third term of the geometric series exceeds the third term of the arithmetric series by 3. The arithmetric series has a positive common difference and the sum of its first three terms equals 54 . Find the first term of both series. Find also the common difference of the arithmetric of the arithmetric series and the common ration of the geometric series.

whyyie

回复 2782# hongji

(5C0(1)^0(-2√3)^5 = - (5C0(1)^0(2√3)^5
5C1(1)^1(-2√3)^4  = 5C1(1)^1(2√3)^4
....

如此推类..

whyyie

equation ax^2 + bx + c =0 with a,b and c as non-zero constants, has roots α and β.

1)if α=入β ,with 入=/= 0 and 入=/=1 ,and equation lx^2 +mx + n =0 with l,m and n as non-zero constants , has roots 入(入+1)^2 and (入+1)^2 .show that [(m^2)/ (ln)] = [(b^2 )/ (ac)]

hongji 发表于 21-9-2011 02:13 PM

ax^2 + bx + c = 0
x^2 + (b/a)x + (c/a) = 0
入β + β = -b/a
入β^2 = c/a
b^2 / ac = (入+1)^2 / 入


lx^2 + mx + n = 0
x^2 + (m/l)x + (n/l) = 0
入(入+1)^2  + (入+1)^2 = -m/l
入(入+1)^2(入+1)^2 = n/l
m^2 / ln = (入+1)^2 / 入

=> [(m^2)/ (ln)] = [(b^2 )/ (ac)]

hongji

find the sum to infinity [3^2(1-x)^2]  +  [3^3(1-x)^3]  + ....3^r(1-x)^r +...and determine the set of x such that this sum exist.


a=3^2(1-x)^2            r=3(1-x)

sum to infinity = a/(1-r)
                      =[9(1-x)^2] /  [3x -2]    >>是这样吗？

然后。。
I r I < 1
I 3 - 3x I <1

3 - 3x <1                3 - 3x > -1
-3x < -2                   -3x> -4
x> 2/3                        x< 4/3
=>>  2/3 < x < 4/3   >>对吗？

hongji

given 2 parallel line L1 and L2 ,passing through (5 ,0 ) and (-5 ,0) respectively and meets the line 4x + 3y =25 rectively at P and Q . If PQ equals 5 units,find the possible slopes of L1 and L2.

slopes = gradient吗？还是radius?

4x + 3y -25 =0
L1 = [4(5) + 3(0) -25] /  [√16 + 9]
L2 = [4(-5) + 3(0) -25]/ [√16 +9]  >>这样吗？

hongji

show that the point P [a(t^2 +1) / 2t   ,  b(t^2 -1) / 2t ) lies on the curve b^2x^2 - a^2y^2 - a^2b^2

show that the equation of the tangent to the curve at P is bx(1^2 + 1 ) - ay(1^2 -1) = 2abt

This tangent cuts the x-axis at A and the y-axis at B. M as the midpoint of OA and O denotes the origin and H divides BM in the ratio 2 : 1. Find the locus of U when t varies.

Euclidzz

回复 2787# hongji


    Yup, you right.

Euclidzz

回复 2789# hongji


   Page1 http://www.facebook.com/photo.ph ... 38178511&type=1
   Page2 http://www.facebook.com/photo.ph ... pe=1&permPage=1

这个是我个人答案，步骤太长了，我打到很累，所以拍下了。。。. T_T
如有错误，请帮我修改。

Euclidzz

given 2 parallel line L1 and L2 ,passing through (5 ,0 ) and (-5 ,0) respectively and meets the line ...
hongji 发表于 22-9-2011 09:49 AM

http://www.facebook.com/photo.ph ... 38178511&type=1
http://www.facebook.com/photo.ph ... pe=1&permPage=1

如有错误，帮我修改。。。><

hongji

如有错误，帮我修改。。。>
Euclidzz 发表于 25-9-2011 12:19 AM

    真的很长。。我抄了，arigatou><

天空之道

(2+1/√2)^5+(2-1/√2)^5=？？

kelfaru

if A =(x is equal n more than 1 bt less than 6)
B = (x more than 0 bt equal n less than 4)
so A union B = ???

书答案是2,3,4
我做到的是1,2,3,4,5

是书答案错了吗？？？

Euclidzz

回复 2794# 天空之道


   

这个是2分的做法。。。如果5分，需要expand给考官看。。。

Euclidzz

回复 2795# kelfaru


A={1,2,3,4,5}
B={1,2,3,4}
AUB={1,2,3,4,5}
也许是书错吧。。。＞＜
那题还有没有其他keyword?

白羊座aries

这个 , integrate sin x (In x ) dx  要怎么做啊？

cheesit92

-lnx(cosx) + integral ((cosx)/x)dx....学了cosine integral? STPM 题目？ 要用Maclaurin来approximate的吧。。。另请高手帮忙。。

Euclidzz

=_= this one is infinite integral..... Not elementary function....