数学Paper 1讨论专区

evildevil7n

回复  evildevil7n


     介意上传你的trial么？
peaceboy 发表于 20-10-2010 08:17 PM

不介意，可是我没有scanner...
不过。。。那个你还是不要抱太大的希望好，你看到的都是最难的了，其他的都是容易的

芭樂

1）Find the coordinates of c, the point of intersection of the curves
    y=e^x and y = 2+3e^-x
if both curves cut the y-axis at the points A and B, calculate the area bounded by AB and the arcs AC and BC
    (Intersection 我找到，不会找area)

2)the curves y = 3 sin x and y = 4 cos x(0<x<pie/2) intersect at point A and meet the x-axis at the origin 0 and the point B (pie/2 , 0) respectively. Prove that the area enclosed by the arcs OA,AB and the line OB is 2 square units.
(整题不会做）
1）（ln3 , 3) , 2 ln 3 units^2

peaceboy

回复 2102# 芭樂


     e^x  = 2+3e^-x
e^2x -2e^x-3 =0
(e^x -3)(e^x +1 )=0
e^x = 3 , -1(rejected)

e^x = 3
x=ln3

y=e^(ln 3)
  =3

curves cut the y-axis
x=0
y=e^0
  =1
A , (0,1)

x=0 , y=2+3e^0
          =5
B , (0,5)

Area bounded = integrate 2+3e^-x_x from 0 to ln 3  -  integrate e^x_x from 0 to ln 3
                    =[2x - 3e^-x]_0 to ln 3  - [e^x]_0 to ln 3
                   = [(2ln3 -1) - (0-3) ] - [3-1]
                   = 2ln 3

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

3 sin x  = 4 cos x
sin x / cos x = 4/3
tan x = 4/3
x = tan^-1  4/3

area = integrate 3 sin x _x from 0 to tan^-1  4/3  + integrate 4 cos x _ x from tan^-1  4/3 to pi/2
       =[-3 cos x]_0 to tan^-1 4/3 + [4sin x]_tan^-1  4/3 to pi/2
       = [-3(3/5) - (-3)] + [ 4- 4(4/5)]
       =6/5 + 4/5
       = 2

peaceboy

回复 2101# evildevil7n


    有机会就上传吧

芭樂

A chord of the circle x^2+y^2 = r^2 is parallel to the x-axis and of the length 2l. The minor segment cur off by this chord is rotated about x-axis to form a solid of revolution. Prove that its volume is 4/3piel^3

peaceboy

回复 2105# 芭樂


x^2+y^2 = r^2
y^2 = r^2 - x^2

volume of the circle from -l to l
pi integrate y^2 dx_x from -l to l = pi [(r^2)x - (x^3) /3]_x from -l to l
                                               = pi [(r^2)l - (l^3)/3] - [-(r^2)l + (l^3)/3]
                                               =2pi [(r^2)l - (l^3)/3]

the remainder things from a cylinder ,with radius (r^2 - l^2)^(1/2) and hight 2l
volume of cylinder = pi[(r^2 - l^2)^(1/2) ]^2 （2l）
                          =2pi [(r^2)l  - l^3]


volume of chord = 2pi [(r^2)l - (l^3)/3] - 2pi [(r^2)l  - l^3]
                       = 2pi [2l^2/3 ]
                       =(4/3)pi l^3

很久没看到这种题目了

xiang911

回复 2100# evildevil7n


   就和别的trial paper不大同咯XD

Lov瑜瑜4ever

今天mathT1的考试题目。。。直接不会做。。。

Given that when f(x) is divided by (x-1)^3, the remainder is 2x^2+x-1.
Find the remainder when f(x) is divided by (x-1)^2.

Lov瑜瑜4ever

我想到的方法是
f(x)=(x-1)^3+(2x^2+x-1)
     =x^3-x^2+4x-2
然后再用(x-1)^2来除x^3-x^2+4x-2
得到remainder=5x-3
考完了才想到的。。。

peaceboy

我想到的方法是
f(x)=(x-1)^3+(2x^2+x-1)
     =x^3-x^2+4x-2
然后再用(x-1)^2来除x^3-x^2+4x-2
得到re ...
Lov瑜瑜4ever 发表于 22-10-2010 06:18 PM

   偶很喜欢这样写的
f(x)/(x-1)^3 = g(x) + (2x^2+x-1)/(x-1)^3
f(x)/(x-1)^2 = (x-1)g(x) + (2x^2+x-1)/(x-1)^2
                 = (x-1)g(x) + (2x^2+x-1)/(x^2 - 2x +1)
                 = (x-1)g(x) + 2 + (5x-3)/(x^2 - 2x +1)
5x-3<<<remainder

Lov瑜瑜4ever

偶很喜欢这样写的
f(x)/(x-1)^3 = g(x) + (2x^2+x-1)/(x-1)^3
f(x)/(x-1)^2 = (x-1)g(x) + (2x^ ...
peaceboy 发表于 22-10-2010 06:25 PM

= (x-1)g(x) + (2x^2+x-1)/(x^2 - 2x +1)
= (x-1)g(x) + 2 + (5x-3)/(x^2 - 2x +1)
为什么那个2要忽略？
看不懂你的做法。。。

peaceboy

回复 2111# Lov瑜瑜4ever


    14/2^3 有remainder  of 6

所以我们这样写
14/2^3 = 1+ 6/2^3


如果现在我们要知道14/2^2 的remainder

14/2^3 = (1)+ 2/2^3
14/2^2 = 2(1) + 6/2^2
           =2(1) + 1 + 2/2^2
现在我们代入14=f(x)
                 2 = (x-1)
                 (1) = g(x)
                 1=2
所以现在有没有比较清楚？


14/2^2 =2(1) + 1 + 2/2^2

f(x)/(x-1)^2 = (x-1)g(x) + 2 + (5x-3)/(x^2 - 2x +1)


忽略掉2因为2已经是整数

Lov瑜瑜4ever

回复  Lov瑜瑜4ever


    14/2^3 有remainder  of 6

所以我们这样写
14/2^3 = 1+ 6/2^3


如 ...
peaceboy 发表于 22-10-2010 07:05 PM

噢噢噢。。。明白了。。。tq。。。

Allmaths

今天mathT1的考试题目。。。直接不会做。。。

Given that when f(x) is divided by (x-1)^3, the remain ...
Lov瑜瑜4ever 发表于 22-10-2010 04:58 PM

    这样也可以:

f(x)=[(x-1)^3]g(x)+2x^2+x-1
     =[(x-1)^3]g(x)+2(x-1)^2+5x-3
     ={[(x-1)^2][(x-1)g(x)+2]}+5x-3

∴remainder=5x-3


或者:
f(x)=[(x-1)^3]g(x)+2x^2+x-1
f'(x)=[3(x-1)^2]g(x)+[(x-1)^3]g'(x)+4x+1
f(1)=2   ---eq 1
f'(1)=5  ---eq 2

f(x)=[(x-1)^2]h(x)+Ax+B
f'(x)=2(x-1)h(x)+[(x-1)^2]h'(x)+A
f(1)=A+B   ---eq 3
f'(1)=A      ---eq 4

equate eq 1 and eq 3,
2=A+B   ---eq 5
equate eq 2 and eq 4,
A=5      

Sub A=5 into eq 5,
2=5+B
B=-3

∴f(x)=[(x-1)^2]h(x)+Ax+B
   f(x)=[(x-1)^2]h(x)+5x-3

remainder=5x-3

偶喜欢用这两种方式。。。

佩琪

-Express the recurring decimal 5.2313131....in the form of a lowest term.
怎样做啊??怎么答案没有n term?!

-The area of a square is 12cm square.From this square,a new square is construct such that each side is 0.9 times the side of the previous square .The process is continued to construct the next square .This goes on indefinitely.If the next square are added together ,what will the total area of the squares be?

peaceboy

-Express the recurring decimal 5.2313131....in the form of a lowest term.
怎样做啊??怎么答案没有n t ...
佩琪 发表于 22-10-2010 10:52 PM

    Express the recurring decimal 5.2313131....in the form of a lowest term.
换成fraction的意思
5.2 +( 0.031 + 0.00031 + 0.0000031 +.....)
52/10 + ( 0.031 + 0.00031 + 0.0000031 +.....)

( 0.031 + 0.00031 + 0.0000031 +.....)
a= 0.031 , r=0.01
sum to infinity的formula，自己试看

sum to infinity formula , a/(1-r)
0.031/(1-0.01)=0.031/0.99
                    = 31/990

52/10+31/990 = 5719/990

the area of a square is 12cm square.From this square,a new square isconstruct such that each side is 0.9 times the side of the previoussquare .The process is continued to construct the next square .Thisgoes on indefinitely.If the next square are added together ,what willthe total area of the squares be?

a=12 , r = 0.9^2 = 0.81
sum to infinity  ,自己试看

12/(1-0.81) = 12/0.19
                 = 1200/19
                 =63.158

wuhu

1){a&#1110;}and {bi}are two A.P. Show that
(a){ai+bi}is also an A.P
(b){1000-ai}is also an A.P
(c)If the sum ∑   ai=A上面n,下面是i =a    and ∑bi=B 上面是n,下面是i=1 ,what  is ∑ 3 ai+5bi 上面是n ,下面是i=1

麻烦各位下~我看不懂题目也不会做~

Allmaths

1){a&#1110;}and {bi}are two A.P. Show that
(a){ai+bi}is also an A.P
(b){1000-ai}is also an A.P
(c ...
wuhu 发表于 23-10-2010 12:05 PM

1a)a1,a2,a3,...,ai<---A.P. with common difference d1  ---eq 1
  b)b1,b2,b3,...,bi<---A.P with common difference d2   ---eq 2


eq 1+eq2,
a1+b1, a2+b2, a3+b3,...,ai+bi


common difference= (a2+b2)-(a1+b1)
                         =(a2-a1)+(b2-b1)
                         =d1-d2   (constant)


∴{ai+bi} is an A.P with common difference d1-d2.



1b){1000-ai}=(1000-a1),(1000-a2),(1000-a3),...,(1000-ai)


common difference=1000-a2-(1000-a1)
                          =a1-a2
                          =-d1
∴{1000-ai} is an A.P with common difference -d1.



1c)∑(3 ai+5bi)=3∑ai+5∑bi

                    =3n+5B






check check 下。。。不知道答案对不对...

evildevil7n

(1+x)^n = co+c1x+c2x^2+c3x^3+c4x^4+...+cnx^n
(a) Let x=1,


(1+1)^n = co+c1 (1)+c2(1)^2+c3 ...
Allmaths 发表于 19-10-2010 09:53 PM

b(i)用nCr=n!/[(n-r)!r!]来prove
  (ii)老师说要用binomial expansion，不然没有分

peaceboy

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

    bi ) 多数人会选择用formula prove的说

b ii 我看没几个人对吧