【纪念当年的帖子（2010）】Add Maths功课讨论区

lin96

还有一个问题，form4 chapter 2 quadratic equation 的，我不知道什么时候要用到 sum of root , product of root, b^2-4ac 这些formula....还有comparison 在什么时候才可以用，因为有时用了对，有时错....

Allmaths

simplifying algebraic expressions by using laws of indices

show that:
25^x-1 + 5^2x+1 = 5^2(x-1) ...
lin96 发表于 4-4-2012 09:15 PM

    请放上挂号。。不然没有人知道你的equation长什么样子。。。

Allmaths

simplifying algebraic expressions by using laws of indices

show that:
25^x-1 + 5^2x+1 = 5^2(x-1) ...
lin96 发表于 4-4-2012 09:15 PM

    花了一点点时间终于知道你写什么了。。。

25^(x-1)+5^(2x+1)=5^(2x-2)+[5^(2x-2)](5^3)
                               =[5^(2x-2)](1+5^3)
                               =(126){5^[2(x-1)]}

josser

The sum of the first hundred terms of an  ap with first term a,  and common different d is T. The sum of the first odd-numbered terms is T/2-1000. Find the value of d.

Allmaths

The sum of the first hundred terms of an  ap with first term a,  and common different d is T. The su ...
josser 发表于 8-4-2012 01:11 AM

    S_100=(100/2)(2a+99d)=T   ---eq 1

T1+T3+T5+...+T99=(T/2)-100
a+a+2d+a+4d+...+a+98d=(T/2)-100
50a+2d(1+2+3+...+49)=(T/2-100)
50a+2d(1225)=(T/2)-100
50a+2450d=(T/2)-100   ---eq 2

sub eq 1 into eq 2,
50a+2450d=[ (100/2)(2a+99d) /2]-100

∴d=4

josser

回复 945# Allmaths


    S_100=(100/2)(2a+99d)=T   ---eq 1

T1+T3+T5+...+T99=(T/2)-100
a+a+2d+a+4d+...+a+98d=(T/2)-100
50a+2d(1+2+3+...+49)=(T/2-100)
50a+2d(1225)=(T/2)-100
50a+2450d=(T/2)-100   ---eq 2

sub eq 1 into eq 2,
50a+2450d=[ (100/2)(2a+99d) /2]-100

∴d=4

我不明白为什么那边是2d

josser

回复 945# Allmaths


    oh, 明白了。。
我之前是跟formula 的，只是拿到 50a + 1225d ...

josser

那么 1225 如何找到？

josser

A GP has non-zero first term a and common ratio r, where 0<r<1. Given that the sum of the 8 terms of the progression is equal to half the sum of infinity. Find the value r , correct to 3 decimal places. Given also that the 17th term of the progression is 10, find a.

Allmaths

A GP has non-zero first term a and common ratio r, where 0
josser 发表于 9-4-2012 08:07 PM

a(1-r^8)/(1-r)=(1/2)[a/(1-r)]r^8=1/2
r= (1/2)^(1/8) or -(1/2)^(1/8)

but 0<r<1, r=(1/2)^(1/8)

ar^16=10
a(1/2)^2=10
a=40

josser

a curvehas equation y=(x-3)^-1 +x
find the coordinates of the maximum point A and the minimum point B on the curve

josser

g(x) = x - 2
f(x) = x^2 - 4x +7
the function h is such that f=hg, and the domain of h is x>0
Obtain an expression for h(x)

josser

the equation of a curve is y = 9/(2-x)
find the volume obtained when the region bounded by the curve, the coordinate axes and the line x=1 is rotated through 360degree about the x-axis.

lin96

想请问第四题的怎么做？

lin96

Given that the quadratic equation mx^2 - 2mx - 1=6x - m has two different roots,
find the range of values of m

Ans: mx^2 - 2mx- 1 - 6x +m =0
       mx^2 - 2mx - 6x - 1 +m=0

b^2 - 4ac > 0
(-2m-6)^2 - 4(m)(-1+m)>0
4m^2 + 24m + 36 + 4m - 4m^2 >0
28m+36>0
28m>-36
m> -36/28
m> -9/7

这个答案对吗？？

leenyiam

a curvehas equation y=(x-3)^-1 +x
find the coordinates of the maximum point A and the minimum point ...
josser 发表于 1-5-2012 09:42 AM

我找到maximum一个而已……这个题目怪怪的……还是我算错了。。

y = (x-3)^-1 + x
dy/dx =  -1(x-3)(1) + 1
          =  -x+3+1
          =  -x+4
turning point, dy/dx = 0
-x+4 = 0
x = 4
sub x=4 into eqn.
y = (4-3)^-1 + 4
y = 5

to determine whether max or min
d^2y/dx^2 = -1 < 0 (max)

maximum point = (4,5)

g(x) = x - 2
f(x) = x^2 - 4x +7
the function h is such that f=hg, and the domain of h is x>0
Obtain an expression for h(x)

f = hg
x^2 - 4x + 7 = h (x-2)
Let x-2 = y
        x   = y + 2
h(y) = (y+2)^2 - 4(y+2) +7
       = y^2 + 4y + 4 - 4y - 8 + 7
       = y^2 + 3
h(x) = x^2 + 3

实话说，我也不懂题目在讲什么。。错的话请通知。谢谢。

ralywong

回复 951# josser


   

a curvehas equation y=(x-3)^-1 +x find the coordinates of the maximum point A and the minimum point B on the curve

y=(x-3)^-1 + x
dy/dx
= -1[(x-3)^-2](1) + 1
=1-(x-3)^-2

For max. or min. point, dy/dx =0
1-(x-3)^-2 =0
(x-3)^-2 = 1
(x-3)^2 = 1/1 =1
x^2 -6x +9 = 1
x^2 - 6x + 8 = 0
(x-4)(x-2)=0
x=4 or x=2

when x=4,
y=(4-3)^-1 + 4 = 1+4 = 5
d2y/dx2
= 2(x-3)^-3
= 2(4-3)^-3
= 2>0  
hence, min. point A = (4,5)

when x =2
y=(2-3)^-1 + 2 = 1
d2y/dx2
= 2(x-3)^-3
= 2(2-3)^-3
=-2<0
hence, max. point = (2,1)

ralywong

回复 954# lin96


   Simplify  [log_81 (2^2) x log_4 (125)]/[log_9 (5)][log_81 (2^2) x log_4 (125)]/[log_9 (5)]
={[(log 4)/(log 81)] x [(log 125)/(log 4)]}/ [log_ 9 (5)]
log 4 可以 cancel 掉
= [(log 125)/(log 81)]/[(log 5)/(log 9)]
= [(log 125)/(log 81)] x [(log 9)/(log 5)]
= [3(log 5)/(log 81)] x [(log 9)/(log 5)]
= [3/(log 81)] x (log 9)
= {3/[2(log 9)]} x (log 9)
= 3/2


其实当你换base的时候不需要特地去选一个号码给它，最好是base 10， 然后simplify完了再 set base

ralywong

回复 952# josser


   [tr][/tr]

g(x) = x - 2 f(x) = x^2 - 4x +7 the function h is such that f=hg, and the domain of h is x>0 Obtain an expression for h(x)

这题我觉得这样写会比较好:f=hg
f(x)=hg(x)
fg^-1 (x) = hgg^-1 (x)
h(x) = fg^-1 (x)
剩下的大致上上几楼的已经接出来了

lin96

回复  lin96


   Simplify  [log_81 (2^2) x log_4 (125)]/[log_9 (5)][log_81 (2^2) x log_4 (125)]/ ...
ralywong 发表于 8-5-2012 02:53 PM

谢谢你！，原来这么简单，可是自己做起来却好复杂