数学Paper 1讨论专区

hongji

deduce the sum of the infinite series 1/(1 x 3)  +  1/(2 x 4)  + 1/(3 x 5)  +...+ 1/ [(n-1)(n+1)] +...


这个怎样做的阿？

Allmaths

[1 / (3-x) ]  <  1 / (x-2) ]
solve.

...我做到。。。
2x - 5 /[(3-x)(x-2)]

然后x< 2 , 5/2
hongji 发表于 15-9-2011 01:49 PM

                 1/(3-x)<1/(x-2)
     1/(3-x)-1/(x-2)<0

(2x-5)/[(3-x)(x-2)]<0
   (2x-5)(3-x)(x-2)<0


∴{x : 2<x<5/2 or x>3}

Allmaths

deduce the sum of the infinite series 1/(1 x 3)  +  1/(2 x 4)  + 1/(3 x 5)  +...+ 1/ [(n-1)(n+1)] +. ...
hongji 发表于 15-9-2011 01:51 PM

   1/(1 x 3)  +  1/(2 x 4)  + 1/(3 x 5)  +...=(1/2)Σ{1/[r(r+2)]}
                                                             =(!/2)Σ[(1/r)-1/(r+2)]

                                                             =(1/2)Σ[(1/r)-1/(r+1)+1/(r+1)-1/(r+2)]
                                                             =(1/2)[f(1)-f(n+1)+f(2)-f(n+2)]
                                                             =[n(3n+5)]/[4(n+1)(n+2)]

hongji

expand (1 + 3x) ^1/3 in ascending powers of x,up to and including the term x^3.




在做by substituting a suitable value of x , find an approximate value of (cube roots 8.24)  correct to 5 decimal places..

x 如何找阿？

Allmaths

expand (1 + 3x) ^1/3 in ascending powers of x,up to and including the term x^3.




在做by subs ...
hongji 发表于 15-9-2011 08:16 PM

    就let x=1/100

[1+3(1/100)]^(1/3)=(103/100)^(1/3)
                               =(1/2)(8.24)^(1/3)

hongji

就let x=1/100

[1+3(1/100)]^(1/3)=(103/100)^(1/3)
                               =(1/2) ...
Allmaths 发表于 15-9-2011 11:36 PM

一定是用x=1/100,如果是其他不一样题目的话？

Allmaths

一定是用x=1/100,如果是其他不一样题目的话？
hongji 发表于 16-9-2011 08:53 AM

    如果用x=1/100, 找出来的approximate value 就比较准。。

当然在这题我用x=1/100也是有根据的。。。

别的题目也当然有另外的算法。。不是每次都x=1/100的。。

hongji

show that a^2 + b^2 >= 2ab



if x + y + z = c , show that x^2 + y^2 + z^2 >= 1/3 c^2

Allmaths

show that a^2 + b^2 >= 2ab



if x + y + z = c , show that x^2 + y^2 + z^2 >= 1/3 c^2
hongji 发表于 17-9-2011 10:20 PM

很经典的题目。。。   
(a-b)^2≥ 0
a^2-2ab+b^2≥ 0
a^2+b^2≥ 2ab


From the result above,
x^2+y^2≥ 2xy   ---(1)
x^2+z^2≥ 2xz   ---(2)
y^2+z^2≥ 2yz   ---(3)


(1)+(2)+(3),
2x^2+2y^2+2z^2≥ 2xy+2xz+2yz   ---(4)


Since (x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz,
then 2xy+2xz+2yz=(x+y+z)^2-x^2-y^2-z^2   ---(5)


Sub (5) into (4),
2x^2+2y^2+2z^2≥(x+y+z)^2-x^2-y^2-z^2
3(x^2+y^2+z^2)≥c^2
x^2+y^2+z^2≥(1/3)c^2   (Shown)


其实还有更容易的method, 就是用cauchy-schwarz inequality不过不在syllabus里面。。


By CS inequality,
(1^2+1^2+1^2)(x^2+y^2+z^2)≥[(1)x+(1)y+(1)z]^2
3(x^2+y^2+z^2)≥(x+y+z)^2
(x^2+y^2+z^2)≥(1/3)c^2   (shown)

hongji

很经典的题目。。。   
(a-b)^2≥ 0
a^2-2ab+b^2≥ 0
a^2+b^2≥ 2ab


From the result above ...
Allmaths 发表于 18-9-2011 12:15 AM

    这是stpm pass year 2000=p
    之后，我还有其他passyear 要问，不好意思=p

hongji

find the equation of the tangent to the curve xy=4 at point P(4,1).
Point A is a point on the x-axis such that PA is parallel to the y-axis.Tangent to the curve xy=4 at P meets the y-axis at point B. The straight line passing through B and parallel to the x-axis meets the curve at point Q. Find the coordinates of Q and show that AQ is a tangent to the curve at Q。
Find the coordinates of the point of intersection of the tangent to the curve xy=4 at P and Q.



每当要show intersect要用discriminat b^2 - 4 ac的吗？

Allmaths

这是stpm pass year 2000=p
    之后，我还有其他passyear 要问，不好意思=p
hongji 发表于 18-9-2011 08:38 AM

    没关系。。不耻下问是好事。。

Allmaths

find the equation of the tangent to the curve xy=4 at point P(4,1).
Point A is a point on the x-axi ...
hongji 发表于 18-9-2011 09:04 AM

   xy=4
x(dy/dx)+y=0
dy/dx=-y/x
At P(4 , 1), dy/dx=-1/4

Equation of tangent at P,
y-1=(-1/4)(x-4)
y=(-1/4)x+2

Point A (4 , 0)
Point B, x=0, y=(-1/4)(0)+2
                       y=2
Point B (0 ,2)

Point Q, y=2, x(2)=4
                            x=2
Point Q (2 ,2)

dy/dx=-y/x
At Q(2 , 2), dy/dx=-1

Equation of tangent at Q,
y-2=-1(x-2)
y=-x+4

Point A (4 , 0) satisfied the equation of tangent at Q, hence AQ is a tangent to the curve at Q.

Equation of tangent at P, y=(-1/4)x+2
Equation of tangent at Q, y=-x+4

Equate both equation,
(-1/4)x+2=-x+4
x=8/3
y=-(8/3)+4
y=4/3

Point of intersection : (8/3 , 4/3)

以我的理解， 题目是要我们找point of intersection of straight line P and Q。。


毕竟我英文不好。。莫怪。。。


上数学好像在上英文课。。

hongji

没关系。。不耻下问是好事。。
Allmaths 发表于 18-9-2011 09:26 AM

    那我不客气咯。。。

hongji

simplify  [ (1 + 2√3)^5 ]  -[ (1 - 2√3)^5 ]


这个哦，是不是这样做的。。。
[(5C0(1)^0(2√3)^5  + 5C1(1)^1(2√3)^4  + 5C2(1)^2(2√3)^3  + 5C3(1)^3(2√3)^2  + 5C4(1)^4(2√3)^1  + 5C5(1)^5(2√3)^0 ] - [(5C0(1)^0(-2√3)^5  + 5C1(1)^1(-2√3)^4  + 5C2(1)^2(-2√3)^3  + 5C3(1)^3(-2√3)^2  + 5C4(1)^4(-2√3)^1  + 5C5(1)^5(-2√3)^0 ]

酱的话等于零吗？

hongji

trial math paper ==做到很粗心，明明会的。。。><

hongji

show that for all real values of x , (x^2 + x + 1) / (x +1)  does not lie between -3 and 1


我只会x=/= 1  ><

hongji

function f is defined f(x)=1/x , with x E R and x=/= 0 ,determine the set of values of x so that f(x)>f(x-1)



这个，f(x-1) = 1/(x-1) , x=/=1 ,然后？。。。

whyyie

show that for all real values of x , (x^2 + x + 1) / (x +1)  does not lie between -3 and 1
hongji 发表于 20-9-2011 09:02 PM

Let y = (x^2 + x + 1) / (x +1)
xy + y = x^2 + x + 1
x^2 + (1-y)x + (1-y) = 0

b^2 - 4ac >= 0
(1-y)^2 - 4(1-y) >= 0
y^2 + 2y - 3 >= 0
(y+3)(y-1) >= 0
y <= -3 , y >= 1

whyyie

function f is defined f(x)=1/x , with x E R and x=/= 0 ,determine the set of values of x so that f(x)>f(x-1)
hongji 发表于 20-9-2011 09:05 PM

1/x > 1/(x-1)
(x-1) / [x(x-1)] > x / [x(x-1)]
-1 / [x(x-1)] > 0
Since -1 / [x(x-1)]  is always positive,
x(x-1) < 0
=> 0 < x < 1