数学Paper 1讨论专区

Allmaths

b(i)用nCr=n!/[(n-r)!r!]来prove
  (ii)老师说要用binomial expansion，不然没有分
evildevil7n 发表于 23-10-2010 08:20 PM

    话说那只是skema所提供的答案...

b(i) 其实也是pascal triangle 的特征...
b(ii) prove 的方法不只binomial可以做..我还见过另一套方法...

数学不是死的...

sim91

这个是我学校的考卷。。。不过从第四题开始就有点不会了麻烦大家指教指教

wuhu

1a)a1,a2,a3,...,ai
Allmaths 发表于 23-10-2010 03:58 PM

   我相信应该对啦~哈哈~谢谢你哦~

sim91

a)If y=ax+be^-x and d^2y/dx^2+2dy/dx=1-e^-x,find the values of a and b.这个题目我不太会。。。

peaceboy

回复 2124# sim91


    y=ax+be^-x

dy/dx = a -be^(-x) ----#
be^(-x) = a-dy/dx ----@

d^2y/dx^2 =be^(-x)
sub @
d^2/dx^2 + dy/dx  = a
add # into both side
d^2/dx^2 + 2dy/dx  = a +a-be^(-x)
d^2/dx^2 + 2dy/dx  = 2a - be^(-x)

d^2y/dx^2+2dy/dx=1-e^(-x)

by comparision , a=1/2
b=1

sim91

回复 2125# peaceboy

原来可以那样做的 我明白了
那么这题应该是要如何做呢
可能我做的方法有点问题  
做到一半就停着了


Find the value of d^2y/dx^2 at the point(1,2)on the curve 2x^3+2y^3-9xy=0
{ans:-108/125}

wuhu

麻烦帮我解决下这题~
Find the first 4 terms in the expansion of (2+x)^-3 . By using a suitable substitution ,obtain the value of 1.99^-3 to 4 decimal places.Show that your answer is accurate up to the 4th decimal place.

peaceboy

回复 2126# sim91


    2x^3+2y^3-9xy=0
6x^2 + 6y^2 (dy/dx) -9[(x)(dy/dx) + y] = 0
6x^2 + 6y^2 (dy/dx) -9(x)(dy/dx) - 9y =0
x=1 , y=2
6+ 24(dy/dx) -9(dy/dx) - 18=0
dy/dx = 12/15

12x + 6[y^2 [d^2y/dx^2] + dy/dx(2y)(dy/dx)] -9x(d^2y/dx^2) - 9dy/dx - 9dy/dx =0
12(1) + 6[4[d^2y/dx^2] + (4)(12/15)(12/15)] -9(d^2y/dx^2) - 9(12/15) - 9(12/15) =0
6[4[d^2y/dx^2] + (4)(12/15)(12/15)] -9(d^2y/dx^2) = 12/5
24[d^2y/dx^2] - 9 (d^2y/dx^2) = 12/5 -384/25
15(d^2y/dx^2) = -324/25
d^2y/dx^2 = -108/125

peaceboy

麻烦帮我解决下这题~
Find the first 4 terms in the expansion of (2+x)^-3 . By using a suitable subst ...
wuhu 发表于 30-10-2010 07:20 PM

    (2+x)^-3 = 2^-3 [1+x/2]^-3
                  =1/8  [1+x/2]^-3
                  = 1/8 [1+ (-3)/1 (x/2) + (-3)(-4)/2! (x/2)^2 + (-3)(-4)(-5)/3! (x/2)^3 +.... ]
                 = 1/8  -3x/16 + 3x^2/16  -5x^3/32

1.99^-3 =[ 2+(-1/100)]^-3
             =  1/8  -3(-1/100)/16 + 3(-1/100)^2/16  -5(-1/100)^3/32
             = 1/8 +3/1600 -3/160000 +5/32000000
             = 0.1269

wuhu

(2+x)^-3 = 2^-3 [1+x/2]^-3
                  =1/8  [1+x/2]^-3
                  = 1/8 [1 ...
peaceboy 发表于 30-10-2010 07:44 PM

   谢谢你~

佩琪

Express [19x^2+40x+9]/ x(x+3)(2x+1) as a sum of simple fractions of lowest term.

佩琪

If (3+kx)^m = 1/9    -    4x/27  +   4x^2/27   - .... ,what are the values oj k and m?For what values of x will series converge?

佩琪

(a) Evaluate 40Mk=18  (k^3-5)
[这一个40和k=18分别在上下，M是转移去左边..]

peaceboy

回复 2131# 佩琪


  Let  [19x^2+40x+9]/ x(x+3)(2x+1) = a/x + b/(x+3) + c/(2x+1)
                                                    = [a(x+3)(2x+1) + bx(2x+1) + cx(x+3)]/x(x+3)(2x+1)

by comparison ,
a(x+3)(2x+1) + bx(2x+1) + cx(x+3) =19x^2+40x+9
let x=0 , 3a = 9
               a=3
let x= -3 , 15b = 60
                  b=4
let x= -1/2 , -5c/4 = -25/4
                      c=5

[19x^2+40x+9]/ x(x+3)(2x+1) = a/x + b/(x+3) + c/(2x+1)
                                           = 3/x + 4/(x+3) + 5/(2x+1)

peaceboy

回复 2132# 佩琪


    (3+kx)^m = 1/9    -    4x/27  +   4x^2/27   - ....

  (3+kx)^m = 3^m (1+kx/3)^m
                 =3^m [1 + m/1!  (kx/3) + m(m-1)/2! (kx/3)^2 +.....]
                 = 3^m + (3^m)m(kx/3)   +(3^m)(m)(m-1)/2  (kx/3)^2+....

by comparison , 3^m = 1/9
                              = 3^(-2)
                           m=-2

(3^m)m(k/3) = -4/27
-2k/27 = -4/27
k=2

for the series to be valid ,      
    l2x/3l<1
   -1  <2x/3< 1
          -3/2 <x < 3/2

peaceboy

(a) Evaluate 40Mk=18  (k^3-5)
[这一个40和k=18分别在上下，M是转移去左边..]
佩琪 发表于 31-10-2010 01:45 PM

    sum of (k^3 -5) , k from 18 to 40  = [ ∑k^3 - ∑5 , k from 1 to 40 ] - [  ∑k^3 - ∑5 , k from 1 to 17 ]
                                                    = [ [(40/2)(1+40)]^2 - 5(40) ] - [ [(17/2)(17+1)]^2 - 5(17) ]
                                                   = 672200 -23324
                                                    = 648876

whyyie

peaceboy

回复 2137# whyyie


    mPQ = [c/p - c/q]/[cp-cq]
        = [(cq-cp)/pq] /[cp-cq]
         = -1/pq
equation of PQ , y-c/p = (-1/pq) (x-cp)
                         y= -x/pq + c/q + c/p

xy=c^2
x(dy/dx) + y =0
dy/dx = -y/x
tangent at P , dy/dx = -(c/p) / cp
                              = -1/p^2
y-c/p = (-1/p^2) (x-cp)
         =-x/p^2 +c/p
y= -x/p^2 +2c/p

===========================================

   tangent at P , dy/dx = -(c/p) / cp
                              = -1/p^2
perpendicular dy/dx = p^2
from the origin , (0,0) equation , y-0 = p^2(x-0)
y=p^2x

intersect from the perpendicular line to the curve
y=p^2x ---1
xy=c^2 ---2
sub 1 into 2
p^2x^2 = c^2
x = + - c/p
y= +- cp
intersect pnt  R=(c/p , cp ),S (-c/p , -cp)
                  


line PR
gradient m1= [c/p-cp]/[cp-c/p]
             =-1
line PS
gradient m2 = (cp+c/p)/[c/p+cp]
                  =1
m1m2 = -1
therefore  PRS is right angle
===============================

R=(c/p , cp ),S (-c/p , -cp)
m = [cp+cp]/[c/p+c/p]
     = 2cp / [2c/p]
    =p^2
y+cp = p^2(x+c/p)
         =p^2x +cp
y=p^2x

=========================
R=(c/p , cp ),S (-c/p , -cp) , Q, (cq , c/q)
mRQ = (cp-c/q)/(c/p - cq)
        = c(p-1/q) / c(1/p - q)
         =[(pq-1)/q] / [1-pq]/p
          =-p/q
,QS = (c/q+cp) /(cq+c/p)
       = (1/q+p) / (q+1/p)
       =[(1+pq)/q]  /[pq+1]/p
        = p/q
(p/q)(-p/q) = -1
p^2/q^2 =1
p^2 = q^2
p= + -q
if P and Q are not the same point , then
p=-q
p+q=0 (shown)

hongji

1)solve the equation  l x l  -  l x - 5 l =5

2)solve the equation l 2x + 7 l - l 6 - 3x l =8

3)solve the equation l 3x +5 l = 2x + 6

Allmaths

1)solve the equation  l x l  -  l x - 5 l =5

2)solve the equation l 2x + 7 l - l 6 - 3x l =8

3 ...
hongji 发表于 1-11-2010 08:43 PM

基本上这些题目都要用graph...比较安全...


Q1:l x l  -  l x - 5 l =5





∴{x : x≥5, xεReal}


Q2: l 2x + 7 l - l 6 - 3x l =8




5x+1=8               or               -x+13=8
      x=7/5                                    x=5

∴x=5,7/5


Q3:  l 3x +5 l = 2x + 6
       (3x+5)^2=(2x+6)^2
9x^2+30x+25=4x^2+24x+36
   5x^2+6x-11=0
  (x-1)(5x+11)=0


∴x=1,-11/5