数学Paper 1讨论专区

peaceboy

1.Find the equations of the circles with radius 5 units, which touch the y-axis and
   pass throug ...
blazex 发表于 5-12-2010 12:59 PM

    Find the equations of the circles with radius 5 units, which touch the y-axis and
   pass through the point (1 , 3 ).

(x-a)^2 + (y-b)^2 =r^2

x=1 , y=3

(1-a)^2 + (3-b)^2 = 25
1 - 2a + a^2 + 9 - 6b +b^2 =25
a^2 -2a +b^2 -6b -15 =0 ---1

x=0 , y=b
(0-a)^2 + (b-b)^2 =25
a^2 =25
a = +-5

from 1
a=5
5^2 -10 + b^2 -6b -15 =0
b^2-6b =0
b= 0,6

a= -5
25+10+b^2 -6b -15 =0
b^2 -6b +20 =0
b^2 -4ac = 36 -80
              = -44 <0 , no real root

equation = (x-5)^2 + (y)^2 =5^2
or (x-5)^2 + (y-6)^2 =5^2


2.Prove that the equation of the chord  PQ where P(ap^2,2ap) and Q (aq^2,2aq)
   lie on the parabola y^2=4ax is (p+q)y-2x+2apq
   Deduce the equation of the tangent at the point at the point P.

P(ap^2,2ap) and Q (aq^2,2aq)
m = [2aq - 2ap] /[ aq^2 - ap^2 ]
    = 2a[q-p] / a[q^2-p^2]
     =2[q-p] / [(q+p)(q-p)]
     = 2/(q+p)

y- 2ap = [2/(p+q)](x-ap^2)
(p+q)[y-2ap] = 2x - 2ap^2
(p+q)y-2(p+q)(ap) = 2x-2ap^2
(p+q)y-2ap^2 - 2apq = 2x-2ap^2
(p+q)y-2x = 2apq

tangent of p ,
ap^2 = aq^2 , 2ap= 2aq
p=q
(p+p)y - 2x = 2ap^2
2py -2x = 2ap^2
py - x = ap^2

hongji

11)find the sum of the positive integers which are less than 150 ad are not multiples of 5 or 7.

multipes by 5:5,10,15...145         multipes by 7:7,14,....140
                    145=5+(n-1)5                          140=7+(n-1)7
                    145=5+5n-5                             140=7+7n-7
                      n=29                                         n=20
                  S5=29/2(5+145)                           S7=20/2(7+140)
                      =2175                                          =1470

怎样找sum of positive integers?

nkrealman

for the series 1,2,3,4,.....149
S149=11175

for the series 35,70,105,140 (intersect)
Sum=350

the sum less than 150 and not the multiple of 7 and 5 =
11175-2175-1470-350=7180

blazex

回复 2283# nkrealman

不是应该+350吗？

hongji

for the series 1,2,3,4,.....149
S149=11175

for the series 35,70,105,140 (intersect)
Sum=350

...
nkrealman 发表于 5-12-2010 04:26 PM

    but answer is 7733
book is wrong?

hongji

回复  nkrealman

不是应该+350吗？
blazex 发表于 5-12-2010 04:43 PM

    for the series 35,70,105,140 (intersect)
Sum=350

为什么这个出现啊？

Allmaths

11)find the sum of the positive integers which are less than 150 ad are not multiples of 5 or 7.

...
hongji 发表于 5-12-2010 03:51 PM

  
Sum of the series positive integers less than 150=1+2+3+4+...+149
                                                             S149=11175

Sum of multiples of 5 less than 150=5+10+15+...+145
                                                 =5(1+2+3+...+29)
                                                 =2175

Sum of multiples of 7 less than 150=7+14+21+...+147
                                                 =7(1+2+3+...+21)
                                                 =1617

Sum of multiples of 5 and 7 less than 150=(5x7)+2(5x7)+3(5x7)+4(5x7)
                                                          =35(1+2+3+4)
                                                          =350

Sum of the positive integers which are less than 150 and are not multiples of 5 or 7
=11175-2175-1617+350
=7733

peaceboy

回复 2286# hongji


    AuB = A + B - (A n B)
5U7 = 5 + 7 - (5 n 7)

所以not multiply of 5 and 7 = All positive integer -[  multiply of 5 +   multiply of 7 -   multiply of (5 n 7)]
                                      = All positive integer -  multiply of 5 -   multiply of 7 + multiply of (5 n 7)

nkrealman

不好意思，是+350没有错

hongji

没想到跟law有关系
明白了。。
我还有另一个问题
G.P的
find the sum of the first ten terms of a G.P .which has 4th term 40 and 9th term 1280.

我只是知道S=a+ar^9=?

peaceboy

没想到跟law有关系
明白了。。
我还有另一个问题
G.P的
find the sum of the first ten terms of a G.P ...
hongji 发表于 5-12-2010 11:27 PM

find the sum of the first ten terms of a G.P .which has 4th term 40 and 9th term 1280.

ar^3 = 40
ar^8 = 1280

找a 找r , 然后用sum 的formula  a[r^n-1]/(r-1)

hongji

find the sum of the first ten terms of a G.P .which has 4th term 40 and 9th term 1280.

ar^3 ...
peaceboy 发表于 6-12-2010 12:13 AM

我有用这方法找过但我比较会用A.P的
40=[a(r^3)]/r-1
1280=[a(r^8-1)]/r-1

40(r-1)=a(r^3-1)
1280(r-1)=a(r^8-1)
divide
32-32=r^-5
r=?

hongji

关于一个问题我不知道我做得对不对
log_12 4 + [1/2log_36 2 + log_36 3]

=[log_2 4/log_2 12]  + [1/2(log_2 2/log_2 36)  +  [log_2 3/log_2 36]


=[2/2+log_2 3] + [1/2(1/2+2log_2 3 )  +[log_2 3/2+2log_2 3]

let u=log_2 3
[2/2+u] +[1/(2/2+u)] +[u/2+2u]

4+4+4u+u^2+2u/4u^2+12u+8=0

u^2+6u+8=4u^2+12u+8
3u^2+6u=0
u(3u+6)=0
u=0 ,u=2
是这样吗？

hongji

u=-2+5i and uv =14+23i

find v in the form a+ib,where a,bE R

peaceboy

回复 2293# hongji


   
ar^3 = 40   ---1
ar^8 = 1280 ---2

2/1 ,
r^5 = 32
      =2^5
r=2

from 1 ar^3 =40
         8a =40
          a=5

sum of the first ten terms = 5[2^10 - 1] / (2-1)
                                     = 5115

peaceboy

回复 2294# hongji


    u=-2+5i and uv =14+23i

let v = a+bi

uv = (-2+5i)(a+bi)
    = -2a + -2bi + 5ai - 5b
    = -2a -5b +(5a-2b)i
uv = 14 + 23i

by comparison
-2a-5b = 14 ---1
5a-2b = 23 ---2

然后自己来

Allmaths

u=-2+5i and uv =14+23i

find v in the form a+ib,where a,bE R
hongji 发表于 6-12-2010 04:02 PM

   
另外一个更快的方法:

uv =14+23i
v=(14+23i)/u
v=(14+23i)/(-2+5i)

rationalize就可以了。。。

hongji

回复  hongji


    u=-2+5i and uv =14+23i

let v = a+bi

uv = (-2+5i)(a+bi)
    = -2a + -2b ...
peaceboy 发表于 6-12-2010 04:42 PM

    找到啦

hongji

另外一个更快的方法:

uv =14+23i
v=(14+23i)/u
v=(14+23i)/(-2+5i)

rationalize就可以 ...
Allmaths 发表于 6-12-2010 05:18 PM

    原来容易的

hongji

4)the quadratic equation x^2+ax +b=0 and x^2+bx+a=0 have a common root .Show that either a=b or a+b+ ...
Log 发表于 3-12-2010 09:33 AM

    有一题我也跟着他方法但做不到
  3x^2+px+1=0and2x^2+qx+1=0have a common root.
  show that 2p^2+3q^2-5pq+1=0

我只作到2p^2+1=0