数学Paper 1讨论专区

harry_lim

去年我不懂, 不过记得今年年头到六月都是walrein一人独行天下.

大家都很厉害, 来这里越多, 发觉自己 ...
whyyie 发表于 8-12-2010 09:27 PM

我不是天才。。。STPM的數學沒有拿A

peaceboy

回复 2356# whyyie


    xy = c^2
x dy/dx + y = 0
dy/dx = -y/x

cp , c/p
dy/dx = - (c/p)/cp
         = -1/p^2
y- c/p = -1/p^2 (x-cp)
p^2y -cp = -x+cp
x=2cp-p^2y


cq , c/q
dy/dx = -1/q^2
y- c/q = -1/q^2 (x-cq)
q^2y -cq = -x+cq
x= 2cq - q^2y

2cp-p^2y= 2cq - q^2y
q^2y - p^2y = 2cq - 2cp
y = 2c(q-p) /(q-p)(q+p)
   =2c/q+p
q+p = 2c/y

x= q(2c-qy)
= q (2c -q(2c/(p+q))
=q[(2cp+2cq-2cq) /(p+q)]
  = 2cpq/p+q
  = pq[2c/(p+q)]
  =pqy
x/y = pq


cp , c/p
cq , c/q
d^2 = (cp-cq)^2 + (c/p-c/q)^2
     =c^2(p-q)^2 +c^2(1/p-1/q)^2
    =c^2[(p-q)^2+(1/p-1/q)^2]
      =c^2 [(p-q)^2+[(q-p)/(pq)]^2]
     =c^2(p-q)^2[1+1/(pq)^2]
     =c^2(p-q)^2[1+1/p^2q^2]

hongji

回复  hongji


    [1/2.3.4+2/3.4.5+3/4.5.6+...+n/(n+1)(n+2)(n+3)]

n/(n+1)(n+2)(n+3) = A/(n+1 ...
junchung2003 发表于 8-12-2010 04:37 PM

    find the sum of the series...
     =-1/2(n+1)  +  2/n+2  -3/2(n+3) >>1

then  sub r=1,r=2,r=3,....,r=n-2,r=n-1,r=n  into 1

r=1,U1=-1/2.2  +   2/3   -  3/2.4
r=2,U2=-1/2.3  +   2/4   -  3/2.5
r=3,U3=-1/2.4  +   2/5   -  3/2.6
..
..
..
r=n-2,Un-2=-1/2(n-1)  +2/n  -3/2(n+1)
r=n-1,Un-1=-1/2.n      +2/n+1 3/2(n+2)
r=n  ,Un=-1/2(n+1)   + 2/n+2  -3/2(+3)

but answer is not same?

whyyie

回复 2362# peaceboy

p+q = 2c / q 貌似和答案有些出入. 答案给p+q = 2x/q. 是问题给错了吗?

junchung2003

回复 2363# hongji


    答案是

(1/4) - (1/2)*( (2n+3)/(n+2)(n+3) ) ？

junchung2003

回复 2364# whyyie


    大概问题吧，我刚才做不到这个所以没有Post解法

whyyie

The straight line y = m(x-1) intersects the parabola y^2 = 2x at two different points P and Q. Show that the midpoint of PQ is (m^2+1)/m), 1/m).

Hence, find the Cartesian equation of the locus of the midpoint of PQ and sketch the locus.

vUwBuC

请问matrix 有没有可能出 2x2 的？
还有，determinant 是不是一定是 Positive 的？

junchung2003

回复 2367# whyyie


y = m(x-1)
Rearrange
y/m + 1 = x -- (1)

y^2 = 2x  -- (2)

Solve (1) and (2)

y^2 = 2(y/m +1)
m*y^2 - 2y - 2m = 0

a = m , b = -2 , c = -2m

Solve y by using formula,

[-b +/- √ b^2 - 4ac ] / 2a

.
.
.
.
y = [ 1 + √ (1+2m^2) ] / m  or [ 1 - √ (1+2m^2) ] / m

If
y =  [ 1 + √ (1+2m^2) ] / m
x = [ 1 + √ (1+2m^2) ] m^2 +1  ---> Assume this is point P

If
y =  [ 1 - √ (1+2m^2) ] / m
x = [ 1 - √ (1+2m^2) ] m^2 +1  ---> Assume this is point Q

Midpoint of PQ
Coordinate of X
= [ [ 1 + √ (1+2m^2) ] m^2 +1 + [ 1 - √ (1+2m^2) ] m^2 +1 ] / 2
.
.
.
= (m^2+1)/m^2

Coordinate of Y
=[ [ 1 + √ (1+2m^2) ] / m + [ 1 - √ (1+2m^2) ] m^2 +1 ] /2
.
.
.
= 1/m

Coordinate of the midpoint is ( (m^2+1)/m^2 , 1/m )

Let
X = (m^2+1)/m^2
Y = 1/m

Rearrange Y=1/m to m = 1/Y

Substitute m = 1/Y into  X = (m^2+1)/m^2

X = ((1/Y)^2+1)/(1/Y)^2
(1/Y)^2 * X = (1/Y)^2+1

Rearrange

x = 1 + y^2

junchung2003

你看看那个X的coordinate对没有，我觉得有power 2

whyyie

The symmetric difference †, of two sets X and Y is defined by
X † Y = (X-Y)U(Y-X)

Using the laws of set algebra, show that for any two sets A and B,
(A U B) † (A n B)  = A † B

junchung2003

回复 2371# whyyie


    之前那个对吗？

whyyie

回复 2372# junchung2003

应该是对了, 我也是做到下面是X^2 .
y= m(x-1)              y^2 = 2x
y^2 = (m^2)(x-1)^2

(m^2)(x-1)^2 = 2x
...
...
...

junchung2003

回复 2373# whyyie


    最后那个locus的答案是什么？

whyyie

回复 2374# junchung2003

没有答案的. 不过找到 x-y^2 - 1 = 0

peaceboy

回复 2371# whyyie


    The symmetric difference †, of two sets X and Y is defined by
X † Y = (X-Y)U(Y-X)

Using the laws of set algebra, show that for any two sets A and B,
(A U B) † (A n B)  = A † B

(A U B) † (A n B)  = [(a u b) - (a n b)] u [(a n b) - (a u b)]
                         = [(a u b) n (a n b)'] u [(a n b) n (a u b)']
                         = [(a u b) n (a' u b')] u [(a n b) n (a' n b')]
                         =[(a u b) n (a' u b')]
                           = [(a u b) n (a')] u [(a u b) n (b')]
                            = [a n a'] u [b n a'] u [a n b'] u [b n b']
                             =[b n a'] u [a n b']
                              = (b-a)u (a-b)
                                = a † b

peaceboy

回复 2368# vUwBuC


    不一定
请到 http://cforum.cari.com.my/viewthread.php?tid=2084079&;extra=page%3D2
学计算机算matrix - -
见7楼的图

peaceboy

最后祝大家好运

FZTeng

今天的Paper，呵呵！
我班11个人，至少一半人不会做！
呵呵！

hongji

回复  hongji


    答案是

(1/4) - (1/2)*( (2n+3)/(n+2)(n+3) ) ？
junchung2003 发表于 8-12-2010 11:44 PM

    answer is n(n+1)/4(N+2)(N+3)