【纪念当年的帖子（2010）】Add Maths功课讨论区

永遠愛著許瑋倫

考着試，所以很久沒上來了
1.A function g is definied by g:x-->a/(b-x) , b is not equal x ,The values 3 and 5 are mapped onto themselves under g.
a.Find the value of a and b
b.State the value of x such that the function g is undefined

harry_lim

考着試，所以很久沒上來了
1.A function g is definied by g:x-->a/(b-x) , b is not equal x ,The values 3 and 5 are mapped onto themselves under g.
a.Find the value of a and b
b.State the value of x such that the function g is undefined
永遠愛著許瑋倫 发表于 31-5-2010 10:55 AM

a. let g(x) = a/(b-x)

    g(3) = 3
    a/(b-3) = 3
        a = 3b - 9 -----------------(1)


   g(5) = 5
   a/(b-5) = 5
   a = 5b - 25 --------------------(2)

   subst. (1) into (2),

   3b - 9 = 5b - 25
   3b - 5b = -25 + 9
   -2b = -16
      b = 8 ............(3)

   subst. (3) into (1)
   a = 3(8) - 9
   a = 24 - 9
   a = 15

   Answer : a = 15 and b = 8  


b. g : x ----> 15/(8-x)  (我不知道這題是不是這樣做的。。。因為我忘記了)

   8 - x =/= 0
     -x =/= -8
      x =/= 8

   the value of x is 8

Enceladus

回复 579# 乙劍真人


   后天要考高数了，今天来临时抱佛脚，遇到不会的题目，请帮帮我。
Given that x=-3 is the equation of the axis of symmetry of the graph of the quadratic function g(x)=x^2 + 2px + 2p - 1, find
(a)the value of p
(b)the minimum value of the function

答案是3和-4.可是我只做到(a)=1......

乙劍真人

回复  乙劍真人

后天要考高数了，今天来临时抱佛脚，遇到不会的题目，请帮帮我。
Given that x=-3 is the equation of the axis of symmetry of the graph of the quadratic function g(x)=x^2 + 2px + 2p - 1, find
(a)the value of p
(b)the minimum value of the function

用 completing the square!

g(x) = x^2 + 2px + 2p - 1
       = x^2 + 2px + (2p/2)^2 - (2p/2)^2 + 2p - 1
       = (x+p)^2 - p^2 + 2p - 1

x + p = 0
-3 + p = 0
       p = 3

minimum value = -p^2 + 2p - 1
                      = -(3)^2 + 2(3) - 1
                      = -4

Enceladus

回复 576# 乙劍真人


   The volume of a sphere increases from 288pi cm^3 to 290pi cm^3.Calculate the approximate increase in its radius.
做到满头大汗

Enceladus

回复 584# 乙劍真人


   谢谢老师的协助，这个月在考试，忙着温习其他科目，有点忽略掉高数了。。。

乙劍真人

回复  乙劍真人
The volume of a sphere increases from 288pi cm^3 to 290pi cm^3.Calculate the approximate increase in its radius.

V = 4/3 pi r^3
dV/dr = 4 pi r^2

delta r = dr/dV x delta V
          = 1/(4 pi r^2) x (290pi - 288pi)       ---(1)

V original = 288pi
4/3 pi r^3 = 288pi
         r^3 = 216
             r = 6 ---------(2)

sub. (2) into (1)
delta r = 1/[4pi (6)^2] x 2pi
          = 2/[4(36)]
          = 1/72 cm

Enceladus

回复 587# 乙劍真人


   为了袮补这次考试的失败，我刚到了大众书局买了高数的书回家去做。先从第一课从新出发，巩固根基。可是刚做几题罢了就出现不会的现象。请老师帮帮我。
The function f is defined by f:x =x^2.
(a)Find the range of values of f for the domain x大过和等于-1,小过和等于4
(b)Determine another domain for f to have the same range

我做错了，连我都不懂错在哪里。。。。。。

Enceladus

回复 587# 乙劍真人


1. A function f is defined by f:x = (a/x) + 2 for all values of x except x = h and a is a constant.(a)Determine the value of h
(b)Given that the value 3 maps to itself under f, find
    (i)the value of a
    (ii)the value of inversed f(-1)

2. f is given by f:x = x/(x-1). Find f^2.

3. Express the quadratic expression (2x^2/3) - (4x/5) -1 in the form a(x+b)^2 +c. Hence, solve the equation (2x^2/3) - (4x/5) -1 = 3. Correct your answer to two decimal places.

4.还有一样，什么是differ by
第一题竟然用0，可以吗?第二题我来来去去都变不出x。第三题是我的死穴，这个我忘记他的做法了。帮帮我。。。。。。

乙劍真人

回复  乙劍真人
为了袮补这次考试的失败，我刚到了大众书局买了高数的书回家去做。先从第一课从新 ...
Enceladus 发表于 5-6-2010 11:56 PM

什么牌子的书？

(a) f(x) = x^2
domain : -1<= x <=4
   x = -1 0 1 2 3 4
f(x) = 1 0 1 4 9 16
then, range = {0 <= f(x) <= 16}

(b) 应该是 0 <= x <= 4

乙劍真人

回复  乙劍真人
1. A function f is defined by f:x = (a/x) + 2 for all values of x except x = h  ...
Enceladus 发表于 6-6-2010 09:26 PM

1. (a) 对，h=0
    (b) (i)[a/(3)] + 2 = 3
                          a = 3
        (ii) f^-1(x) = 3/(x-2)
            f^-1(-1) = -1

2. ff(x) = [x/(x-1)]/{[x/(x-1)] - 1}
           = x/[x-1(x-1)]
           = x

3. 看不懂题目

4. differ 就是 different 的意思咯，请给完整的题目..

Enceladus

回复 591# 乙劍真人


   那本书的封面是褐色和橙色。Pelangi的，作者是Pauline Wong Mee Kiong。题目是SPM Quick Revision Additional Mathematics。图片有一个螺这样的东西。

乙劍真人

回复  乙劍真人
3. Express the quadratic expression (2x^2/3) - (4x/5) -1 in the forma(x+b)^2 +c. Hence, solve the equation (2x^2/3) - (4x/5) -1 = 3.Correct your answer to two decimal places.

Allmaths

各位好!

想请问哪儿有得download SPM add maths past year questions 吗?

ling_wen

请帮帮忙吧>.<(form 4, chapter 1)


1. Given the function g(x)= 4x+7/x-2 , x not equal to 2 , find the objects that are mapped onto themselves.

2. Given the function h:x=x-6/x^2-4, state the values of x such that the function
h is undefined.

3. Given the function g:x=3x^2-5, find the values of t such that
g(t)=-7t/2.

4. Given the functions f:x=p-2x and g:x=3x^2-7, find the value of p if fg(2)=4.

5. Given that g:x=x^2-3, find g^2.

Allmaths

请帮帮忙吧>.
ling_wen 发表于 10-6-2010 08:13 PM

1) obejcts mapped onto themselves, g(x)=x
4x+7/x-2 =x                                                                                            4x+7=x^2-2x
      0=x^2-6x-7
      0= (x-7)(x+1)
  x=7      or      x=-1


2) function undefined, denominator=0

x^2-4=0
(x-2)(x+2)=0
  x=2     or      x=-2


3) g(x)=3x^2-5
    g(t)=3t^2-5=-7t/2
          6t^2-10=-7t
     6t^2+7t-10=0
     (6t-5)((t+2)=0

    t=5/6    or    t=-2


4) g(x)=3x^2-7
    g(2)=3(2)^2-7
          =5

fg(2)=f[g(2)]
       =f(5)
       = p-2(5)
       = p-10

since fg(2)=4
         p-10=4
              p=14

5) g(x)=x^2-3

    g^2(x)= gg(x)
           = g[g(x)]
           = g(x^2-3)
           = (x^2-3)2-3
           = x^4 -6x^2+6

ling_wen

1) obejcts mapped onto themselves, g(x)=x
4x+7/x-2 =x                                         ...
Allmaths 发表于 10-6-2010 09:08 PM

thanks for ur answer~
^^谢谢~

josser

请帮忙 (form 5, chapter 3)

3 points A,B,C are such tat OA= 5i+6j , OB= 6i+4j and OC=ki+2j , where O is
                                                ~  ~           ~   ~              ~  ~  
the origin. find the value of k if the points A, B, C are collinear.

乙劍真人

请帮忙 (form 5, chapter 3)

3 points A,B,C are such tat OA= 5i+6j , OB= 6i+4j and OC=ki+2j , where O is
                                                ~  ~           ~   ~              ~  ~  
the origin. find the value of k if the points A, B, C are collinear.
josser 发表于 14-6-2010 04:52 PM

这个是 chapter 5 噢..

我们必须找 AB 跟 BC (因为这三个点是一个直线)

AB = AO + OB
     = -(5i+6j) + (6i+4j)
     = i-2j
BC = BO + OC
     = -(6i+4j) + (ki+2j)
     = (-6+k)i-2j

Collinear: AB = pBC
             i-2j = p[(-6+k)i-2j]

             1 = p(-6+k) -------(1)
             -2=-2p -------------(2)

From (2), p=1 -----(3)
sub.(3) into (1)
              1= -6+k
             k = 7

josser

回复 599# 乙劍真人


    oh。。。原来是这样～ 谢谢你呀！

还有一题， 同样chapter 的：

given tat a=7i+8j ,  b=2i-5j   and   c=13i-7j , find |a+b+c|
                ~   ~   ~    ~  ~  ~          ~     ~ ~          ~ ~ ~