数学Paper 1讨论专区

Saturn20

需要大家的帮忙。。。

Given that 2^a =3^b=18^cshow that ab=c(b+2a)

谢谢。。。。

Allmaths

需要大家的帮忙。。。

Given that 2^a =3^b=18^cshow that ab=c(b+2a)

谢谢。。。。
Saturn20 发表于 16-6-2011 11:06 AM

    Let x=2^a=3^b=18^c

2=x^(1/a), 3=x^(1/b), 18=x^(1/c)

Since 2x3^2=18,

[x^(1/a)][x^(1/b)]^2=x^(1/c)
             (1/a)+(2/b)=1/c
                  c(b+2a)=ab      (shown)

shin93

erm...刚转校 = = 还没学老师就教完了 TT

hongji

1) solve simultaneous equation
x^2 = 4x - y   and   y^2 = 4y - x




2) (3x^4) + (4x^3) -(14x^2) +4x +3 =0
我做到一半？？
因为U= x-(1/x)
那么[(x+(1/x)] [(x-(1/x)] =?

hongji

sketch the graph of y = I 1-2x I  -1.by using your graph ,draw a suitable straight line to solve inequality I 1-2x I < x+1

yuanhua88

回复 2621# Saturn202^a=3^b
ln(2^a)=In(3^b)
a In2= b ln3
ln3=(a/b)*ln2 -------->(1)

18 = 2*3^2
In 18 = In (2*3^2)
          = ln 2 +2 ln 3

substitute In3 with the 1st equation
ln 18 = ln2 +2*(a/b)*ln 2
          =In 2^(1+a/b)
18= 2^((b+a)/b) ---------> (2)

2^a=18^c
a log2 (2) = c log2 (18)
log2 (2) =1;
a = c log2 (2^((b+a)/b))
   =(c*(b+a)/b) log2 (2)
a =c*(b+a)/b
ab=c*(b+a)

有点乱，希望你能明白

yuanhua88

erm...刚转校 = = 还没学老师就教完了 TT
shin93 发表于 16-6-2011 07:01 PM

1. z^2 = (a+bi)(a+bi)           = (a^2 -b^2)+(2ab)



a^2-b^2=2 ----->(1)

2ab=2*sqrt(3)-->(2)   [注：sqrt=square root]

solve equation (2)
ab=sqrt(3)
a=(sqrt(3))/b --->(3)

Substitute (3) into (1)
((sqrt(3))/b)^2 -b^2=2
3/b^2 -b^2 =2
3-b^4=2b^2
b^4+2b^2-3=0

Let b^2=X
X^2+2X-3=0
X=1 or X=-3
b^2=-3 (invalid) so b^2=1
b=1 or b=-1
a=-sqrt(3) when b=1
a=sqrt(3) when b=-1

z1=-sqrt(3)+i
z2=sqrt(3)-i

水晶雪球

想问一下大家，如果我在spm add maths的成绩是C+（之前从未pass过），那我是不是不太适合拿maths S呢???

Log

回复 2599# blazex
1)solve the equation (sin^-1 X)-(cos^-1 X)=(sin^-1(1-x))

2)without using table or calculators, show that (cos^-1(3/5))-(tan^-1(-3/4))=pi/2.


1. (sin^-1 X)-(cos^-1 X)=(sin^-1(1-x))
    let a= sin^-1 ---> sin a =x
         b= cos^-1 ---> cos b =x            , thus  sin b  = √(1-x&#178; )         
         c= (sin^-1(1-x))---> sin c = 1-x , thus  cos c =√(2x-x&#178; )

(sin^-1 X)-(cos^-1 X)=(sin^-1(1-x))
       a -b =c
       a= c+b
   sin a = sin(c+b) = sin c cos b + cos c sin b
         x = (1-x)(x) +√(2x-x&#178; ) .√(1-x&#178; )                                       
solve and get the answer..
x = 0 , 0.78078, -1.28078 and bcoz  -1≤ sin θ ≤ 1  so  x ≠ -1.28078
  ∴x  = 0 , 0.78078


2. 注： 我是假设  (cos^-1(3/5))  和  tan^-1 ((-3/4)) 的angle 在 -90° 和90°之间.因为问题没给range..

let  m = (cos^-1(3/5)) ---> cos m = 3/5  ,  sin m = 4/5  
       n =  tan^-1 ((-3/4)) ---->  tan n = -3/4  , sin n = -3/5  ,cos n = 4/5

(cos^-1(3/5))-(tan^-1(-3/4)) = m-n
sin 【 (cos^-1(3/5))-(tan^-1(-3/4)) 】 = sin (m-n)
                                                              =  sin m cos n - cos m sin n
                                                               = (4/5)(4/5) -(3/5)(-3/5)
sin 【 (cos^-1(3/5))-(tan^-1(-3/4)) 】   = 1
                ∴   (cos^-1(3/5))-(tan^-1(-3/4))=pi/2.

Log

回复 2624# hongji

1) solve simultaneous equation
x^2 = 4x - y   and   y^2 = 4y - x

2) (3x^4) + (4x^3) -(14x^2) +4x +3 =0


1.) x^2 = 4x - y                            y^2 = 4y - x -------(2)
       y = 4x-x^2 ----- (1)  

insert (1) to (2),
( 4x-x^2)^2 = 4(4x-x^2) -x
x ( x^3 -8x^2 +20x  -15) =0
x =0 ,       or         x^3 -8x^2 +20x  -15=0
                              (x-3)(x^2-5x+5)=0
自己来.......................

2.)  (3x^4) + (4x^3) -(14x^2) +4x +3 =0
建议用   u= x+1/x

Log

回复 2625# hongji

file:///C:/DOCUME%7E1/Toshib@/LOCALS%7E1/Temp/moz-screenshot-2.pngfirst , sketch  y= l 1-2x l -1 = 2x-2 , x>1/2  or  -2x , x </=1/2
file:///C:/DOCUME%7E1/Toshib@/LOCALS%7E1/Temp/moz-screenshot-4.pngfile:///C:/DOCUME%7E1/Toshib@/LOCALS%7E1/Temp/moz-screenshot-5.pnghttp://www.wolframalpha.com/input/?i=abs%28+1-2x%29-1+
then, l1-2xl<x+1
  l1-2xl -1 < x
             y <x
file:///C:/DOCUME%7E1/Toshib@/LOCALS%7E1/Temp/moz-screenshot-6.pngso the line y=x is drawn on the graph  ans u ll find that the line intersect the y = l1-2xl -1 at two points
to find two points(x-coordinate) solve y=2x-2 with y=x    and    y= -2x with y=x simultaneously   
http://www.wolframalpha.com/input/?i=abs%28+1-2x%29-1+%3C+x

from graph , for      l1-2xl<x+1 , the solution of x is {x : 0<x<2}
                       



file:///C:/DOCUME%7E1/Toshib@/LOCALS%7E1/Temp/moz-screenshot-3.png

Wangs

请问怎么prove that   [ { (A ∪ B) ∩ C' }∪ B']'= B ∩ C

Log

回复 2632# Wangs
[ { (A ∪ B) ∩ C' }∪ B']'= B ∩ C

[ { (A ∪ B) ∩ C' }∪ B']'
= { (A ∪ B) ∩ C' }‘ ∩ B              De Morgan's Law
={ (A∪B)' ∪ C }  ∩   B              De Morgan's Law
=  { ( A'∩ B') ∪C} ∩   B               De Morgan's Law
=  { C ∪ ( A'∩ B') } ∩  B              Commutative Law
= { (C ∪A') ∩ (C ∪ B') } ∩ B      Distributive Law
=  (C ∪A') ∩ { (C ∪ B') ∩ B }      Associative Law
= (C ∪A') ∩ {  B ∩ (C ∪ B') }       Commutative Law
=  (C ∪A')∩{ (B ∩ C) ∪(B∩B') }  Distributive Law
=  (C ∪A') ∩ { (B ∩ C) ∪ Φ}        Complementary Law
=   (C ∪A') ∩ (B ∩ C)                      Identity Law
=     (C ∪A') ∩( C ∩ B)                    Commutative Law
=      { (C ∪A') ∩ C } ∩ B                   Associative Law
=  {C ∩ (C ∪A')} ∩ B                     Commutative Law
=    C ∩ B                                        Absorption Law
=    B ∩ C (proven)                          Commutative Law

hongji

回复  hongji

1) solve simultaneous equation
x^2 = 4x - y   and   y^2 = 4y - x

2) (3x^4) + (4 ...
Log 发表于 23-6-2011 03:47 PM

    u=x-1/x 这个用过了，还是不能因为有2个出现，一个(x^2 - [1/x^2])  和 (x +1/x)

hongji

find the equation of the tangent at T (3t^2 , 6t) to the parabola y^2 = 12 x .
If tangent at T meets the line x+3=0 at R , find the coordinates of R.Show that the locus R is a straight line parallel to the y-axis as t as varies

gradien of tangent= 1/t











find the equation of the tangent at P(ct , c/t) to the rectangular hyperbola xy=c^2.
If a line from the origin perpendicular to this tangent meets at Q, prove that OP . OQ =2c^2.
Find the equation of the locus of Q as t varies.

Log

回复 2634# hongji
err....我是说用 u =x +1/x ....

hongji

回复  hongji
err....我是说用 u =x +1/x ....
Log 发表于 26-6-2011 11:41 AM

    哦对，u= x+1/x   所以 u^2=x^2  +  [(1/x^2)]
  那么(x^2 - [1/x^2])  是什么啊

Allmaths

哦对，u= x+1/x   所以 u^2=x^2  +  [(1/x^2)]
  那么(x^2 - [1/x^2])  是什么啊
hongji 发表于 26-6-2011 04:59 PM

    不对啊。。。


u=x+(1/x)

u^2=x^2+2x.(1/x)+(1/x)^2

u^2=x^2+(1/x^2)+2

注意：(a+b)^2=a^2+2ab+b^2

Allmaths

find the equation of the tangent at T (3t^2 , 6t) to the parabola y^2 = 12 x .
If tangent at T meet ...
hongji 发表于 26-6-2011 08:09 AM

    (1)equation of tangent,
     y-6t=(1/t)(x-3t^2)
         ty=x+3t^2

Intesect with x+3=0, x=-3:
ty=(-3)+3t^2
  y=3(t^2-1)/t

∴R(-3 , 3(t^2-1)/t) , locus x=-3 is parallel to y-axis

red_snail

回复 2633# Log


   可以解释一下absorption law吗？没看过