数学Paper 1讨论专区

whyyie

Allmaths

whyyie 发表于 18-11-2010 03:03 PM

   
(1-x)^-2=1+2x+3x^2+4x^3+5x^4+...+(n+1)x^n

∴coefficient of x^n is n+1

expansion valid : |-x|<1
                      -1<x<1
  
Σ (n/2^n) from 0 to infinity=1/2+2/4+3/8+4/16+...
                                      =(1/2)[1+2(1/2)^1+3(1/2)^2+4(1/2)^3+...]
                                      =(1/2)(1-x)^-2      (Let x=1/2)
                                      =(1/2)(4)
                                      =2

whyyie

帮忙一下, 用红色圈的不会做

Part (c) 是 summation of ( r = n+3 to 2n)

Lov瑜瑜4ever

帮忙一下, 用红色圈的不会做

Part (c) 是 summation of ( r = n+3 to 2n)
whyyie 发表于 19-11-2010 03:20 PM

我做C的
r from n+3 to 2n
一共有n-2个terms

Sn=n/2(a1+an)
312=[(n-2)/2](n+3+2n)
(n-2)(3n+3)=624
(n-2)(n+1)=208
n^2+n-2n-2-208=0
n^2-n-210=0
n=15 or n=-14(ignored)
hence, n=15#

Allmaths

帮忙一下, 用红色圈的不会做

Part (c) 是 summation of ( r = n+3 to 2n)
whyyie 发表于 19-11-2010 03:20 PM

    之前在这里做过类似的题目...

Allmaths

帮忙一下, 用红色圈的不会做

Part (c) 是 summation of ( r = n+3 to 2n)
whyyie 发表于 19-11-2010 03:20 PM

    equation of normal, y+px=2ap+ap^3   --->eq 1

Q( aq^2 , 2aq)

sub x=aq^2 and y=2aq into eq 1,

2aq+p(aq^2)=2ap+ap^3
2aq+apq^2=2ap+ap^3
2a(q-p)=-ap(q-p)(p+q)
q=-p-2/p   (shown)




对了whyyie, 你今年有拿FMT吗?

tiewkj

回复 2208# eassaic

老实说，数学没有很难，比起其他科目的话（像化学物理那种），做多多练习，A都有得你拿。加油吧～祝服你～

blazex

1)Find the equation of the tangent at T (3t^2,6t) to the parabola y^2=12X. If the
   tangent at T meets the line X+3= at R. find the coordinates of R. Show that the  
   locus of R is a straight line parallel to the y-axis as t various.
#我得到的equation是y=X/t+3t 而 R的coordinates是（-3，3t-3/t)。只是不懂如何证明。

2）Find the equation of the normal at P(5t^2 , 10^t) to the parabola y^2=20X.
     If the normal at P meets the parabola again at Q, find the coordinates of Q

Allmaths

1)Find the equation of the tangent at T (3t^2,6t) to the parabola y^2=12X. If the
   tangent at T meets the line X+3= at R. find the coordinates of R. Show that the  
   locus of R is a straight line parallel to the y-axis as t various.
blazex 发表于 24-11-2010 09:16 AM

x=-3 已经是parallel to y-axis 了。。。

Allmaths

2）Find the equation of the normal at P(5t^2 , 10^t) to the parabola y^2=20X.
     If the normal at P meets the parabola again at Q, find the coordinates of Q
blazex 发表于 24-11-2010 09:16 AM

   
Equation of normal,
y+tx=5t^3+10t

From original equation,
y^2=20x
x=(y^2)/20
Sub into equation of normal,
y+ty^2/20=5t^3+10t
ty^2+20y-(100t^3+200t)=0

用quadratic formula 后得 y=10t or -(20+10t^2)/t

take y=-(20+10t^2)/t,
x=(1/20)[-(20+10t^2)/t]^2
x=(5/t^2)(t^2+2)^2

∴Q( (5/t^2)(t^2+2)^2   ,   -(20+10t^2)/t   )

hongji

when x^4+px^3+qx^2+rx+6 is divided by x+1 ,x-1 and x+2 , the remainder are 6,12 and 54 respectively. Find the values of p,q, and r



f(-1)=6
f(1)=12
f(-2)=54

(-1)^4+p(-1)^3+q(-1)^2+r(-1)+6=6
7-p+q-r=6
p-q+r=1.............1

(1)^4+p(1)^3+q(1)^2+r(1)+6=12
p+q+r=5.............2

(-2)^4+p(-2)^3+q(-2)^2+r(-2)+6=54
-8p+4q-2r=32........3

2..p=5-q-r
2into1  5-q-r-q+r=1
           q=2

2into3  -8(5-q-r)+4q-2r=32
r=-8

from 1 : p-2-8=1
             p=11


why answer is p=3 q=2 r=0

Allmaths

when x^4+px^3+qx^2+rx+6 is divided by x+1 ,x-1 and x+2 , the remainder are 6,12 and 54 respectively. ...
hongji 发表于 28-11-2010 06:04 PM

    应该是x-2才对...

hongji

find the maximum or minimum value and the corresponding values of x for the following quadratic expression

2+2x-[x^2/2]
i use ..
y=[-x^2/2]+2x+2
y=-x^2+4x+4
i find the maximum value of y is 8 when x=2
answer is x=2 and y=4
i know when x=2  , then y=4
but i don't know the correct method..

hongji

if the equation x^2+3(a+3)x-9a=0 as repeated root,find the possible values of a.

i us b^-4a=0  
x^2+3ax+9x-9a=0

(3a+9)^2 - 4(1)(-9)=0
9a^2+54a+117=0
i use quadratic formula a=-1,-5
but answer is a=-1 or -9

junchung2003

回复 2234# hongji

应该是

(3a+9)^2 - 4(1)(-9a)=0

至于上面那题，是什么chapter的？
请用differentiation

Let f(x) = 2+2x-[x^2/2]
f'(x) = 2 - x

when f'(x) = 0

2 - x = 0
x = 2

f''(x) = -1 (Since it is negative, the turning point is maximum point)

When x = 2, y = 4

Therefore, the maximum value is 4 corresponding to x

hongji

回复  hongji

应该是

(3a+9)^2 - 4(1)(-9a)=0

至于上面那题，是什么chapter的？
请用differenti ...
junchung2003 发表于 28-11-2010 11:55 PM

    这两题是chap2polynomial的
differentiation我还没学
原来看少了9a
谢谢

junchung2003

那么你的老师有给你解法吗？
polynomial好像solve不到Maximum/Minimum的问题列

kelfaru

这两题是chap2polynomial的
differentiation我还没学
原来看少了9a
谢谢
hongji 发表于 29-11-2010 04:13 PM

   
可以用completing the square....

hongji

那么你的老师有给你解法吗？
polynomial好像solve不到Maximum/Minimum的问题列
junchung2003 发表于 29-11-2010 05:50 PM

    我是做longman的参考书的练习
所以老师没教。。
只是有答案

hongji

可以用completing the square....
kelfaru 发表于 29-11-2010 05:53 PM

我有用过completing the square
2+2x-x^2
y=-x^2+4x+4
=-[(x-2)^2-8]
答案是maximum point y is 4
我只是知道whenx=2,y就是4