数学Paper 1讨论专区

nkrealman

回复 2180# Allmaths


   conjugate没有错，expand的时候错了，没关系，反正我也找到了

Allmaths

回复  Allmaths


   conjugate没有错，expand的时候错了，没关系，反正我也找到了
nkrealman 发表于 3-11-2010 02:10 AM

    看到了...我也好奇下我怎么拿到答案...

數學神童

因为你后来square了整个equation，你的mistake被squaring method给cancel了

hongji

2^x-1 = 3^x+1 give answer correct to 2 decimal places..

2^x . 2^-1 = 3^x . 3^1
    then..
     1/2 . 2^x=3 . 3^x  ?

hongji

evaluate x if log 2(base) (1+x) +log 2(base) (5-x)-log2(base) (x-2)=3


who can help me?

Lov瑜瑜4ever

evaluate x if log 2(base) (1+x) +log 2(base) (5-x)-log2(base) (x-2)=3


who can help me?
hongji 发表于 3-11-2010 08:10 PM

log 2(base) (1+x) +log 2(base) (5-x)-log2(base) (x-2)=3
log 2(base) [(1+x)(5-x)/(x-2)] =3
(1+x)(5-x)/(x-2)=2^3
5-x+5x-x^2=8x-16
x^2+4x-21=0
(x+7)(x-3)=0
x=-7 or x=3
x=-7 is rejected as it will make (1+x)<0
so the final answer is x=3

peaceboy

2^x-1 = 3^x+1 give answer correct to 2 decimal places..

2^x . 2^-1 = 3^x . 3^1
    then..
      ...
hongji 发表于 3-11-2010 08:08 PM

   出题目 bracket 很重要
2^(x-1) = 3^(x+1)
(2^x)/2 = 3*3^x
(2/3)^x = 6
log (2/3)^x = log 6
x log (2/3) = log 6
x= log6/log (2/3)

whyyie

Prove that the straight line x -ty+4t^2 = 0 touches the parabola y^2 = 16x for all values of t and find the point of contact

peaceboy

回复 2188# whyyie


   Prove that the straight line x -ty+4t^2 = 0 touches the parabola y^2 = 16x for all values of t and find the point of contact

x -ty+4t^2 = 0
x= ty - 4t^2


y^2 = 16x
y^2 = 16(ty - 4t^2)
      =16ty - 64t^2
y^2 - 16ty +64t^2 =0


b^2-4ac = (-16t)^2 - 4(1)(64t^2)
            =256t^2 - 256t^2
             =0 , >=0
therefore the straight line x -ty+4t^2 = 0 touches the parabola y^2 = 16x for all values of t


intersect point,y =[ -b +-(b^2-4ac)^(1/2) ] /2a
                     =-b/2a
                     = 16t/2
                     =8t
x= ty - 4t^2
  =t(8t) -4t^2
  =4t^2


intersect point = (4t^2 , 8t)

hongji

log 2(base) (1+x) +log 2(base) (5-x)-log2(base) (x-2)=3
log 2(base) [(1+x)(5-x)/(x-2)] =3
(1+x ...
Lov瑜瑜4ever 发表于 3-11-2010 09:57 PM

2^3原来是从log_2来得。。。
还有一题。。
log_2 x - log_2 y = 2
答案是x=4y

是不是也跟这个方法一样。。
[log_2 x / log_2 y]=2
就是从log_2
变成[x/y]=2^2
    [x/y]=4
      x=4y

hongji

出题目 bracket 很重要
2^(x-1) = 3^(x+1)
(2^x)/2 = 3*3^x
(2/3)^x = 6
log (2/3)^x =  ...
peaceboy 发表于 3-11-2010 10:37 PM

    你会[√3/2]  变成  [√6]/2？

lonely_world

你会[√3/2]  变成  [√6]/2？
hongji 发表于 4-11-2010 08:47 PM

[(√3)/(√2)]x[(√2)/(√2)]
=(√6)/2

lonely_world

if y(√x) - √y - x = 1.
Determine the value of y and dy/dx when x=1.

How to differentiate √y ?

Lov瑜瑜4ever

2^3原来是从log_2来得。。。
还有一题。。
log_2 x - log_2 y = 2
答案是x=4y

是不是也跟这个方法 ...
hongji 发表于 4-11-2010 08:43 PM

   嗯嗯

peaceboy

if y(√x) - √y - x = 1.
Determine the value of y and dy/dx when x=1.

How to differentiate √y ?
lonely_world 发表于 4-11-2010 09:42 PM

   d [(y)^1/2] /dx = (1/2)(y)^(-1/2) dy/dx
                        = (1/2y) (dy/dx)

whyyie

Lov瑜瑜4ever

whyyie 发表于 5-11-2010 10:30 AM

(1+3x)^n

for term x^(m+1),   (nCm+1)(3x)^(m+1)=(nCm+1)[3^(m+1)][x^(m+1)]

for term x^m,  (nCm)(3x)^m=(nCm)(3^m)(x^m)

hence, (nCm+1)[3^(m+1)]=3/2[(nCm)(3^m)]
          (3^m)(3)[n!/((m+1)!(n-m-1)!)]=3/2(3^m)[n!/(m!(n-m)!)]
          1/[(m+1)!(n-m-1)!]=1/[2(m!)(n-m)!]
           [m!(n-m)!]/[(m+1)!(n-m-1)!]=1/2
               (n-m)/(m+1)=1/2
               2n-2m=m+1
               2n-3m-1=0  (shown)

lonely_world

請問下
by grouping the terms of the series 1^2 -3^2 + 5^2 -7^2 +... in pairs, or otherwise, find the sum of the first 2n terms of the series. Hence, show that the sum of the first 2n+1 terms is 8n^2 + 8n +1.
這題有人會做嗎?

蠻緊急的~
求求大家!

Lov瑜瑜4ever

請問下
by grouping the terms of the series 1^2 -3^2 + 5^2 -7^2 +... in pairs, or otherwise, find th ...
lonely_world 发表于 9-11-2010 02:15 PM

For sum of 2n terms,
  1^2-3^2+5^2-7^2+...+(4n-3)^2-(4n-1)^2
=(1-3)(1+3)+(5-7)(5+7)+...+(4n-3-4n+1)(4n-3+4n-1)
=(-2)(4)+(-2)(12)+...+(-2)(8n-4)
=(-2)(4+12+...+8n-4)
=(-2)(n/2)(4+8n-4)
=-8n^2

For sum of 2n+1 terms,
  1^2-3^2+5^2-7^2+...+(4n-3)^2-(4n-1)^2+(4n+1)^2
=[1^2-3^2+5^2-7^2+...+(4n-3)^2-(4n-1)^2]+(4n+1)^2
=-8n^2+(4n+1)^2
=-8n^2+16n^2+8n+1
=8n^2+8n+1 (Shown)

peaceboy

回复 2199# Lov瑜瑜4ever


    以后就靠你了