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Author Topic: Pure math(p3)  (Read 7939 times)

Offline candy

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Re: Pure math(p3)
« Reply #30 on: May 28, 2009, 03:55:07 pm »
second part pleasee!

The value of 'a' is 8
« Last Edit: May 28, 2009, 03:56:46 pm by candy »
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Offline sahrhp

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Re: Pure math(p3)
« Reply #31 on: May 28, 2009, 04:39:09 pm »
SEE EARLIER PAGE 2
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Offline louis

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Re: Pure math(p3)
« Reply #32 on: May 28, 2009, 04:44:12 pm »
One way to get the other factor of  (x^4) -2x³-2x² +8  is doing the long division. 
 Divided by x² – 4x + 4 ,  you  get x² + 2x + 2

  Now,  (X^4 )- 2x³-2x² +8  =( x ²– 4x + 4)( x² + 2 x + 2)
                                     =(  x – 2 ) ² {   ( x + 1 ) ² + 1 }
(  x – 2 ) ² is always positive
( x + 1 ) ² + 1 is always positive too.
So (X^4 ) -2x³-2x² +8  cannot be negative
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Offline crucio

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Re: Pure math(p3)
« Reply #33 on: May 28, 2009, 05:25:33 pm »
another rather tricky way of doing it is:

y=x^4 + 2x^3 -2x^2 +8
{\frac{\delta y}{\delta x}}=4x^3 -6x^2 -4x
{\frac{\delta y}{\delta x}}=0 \Rightarrow 4x^3 -6x^2 -4x=0\Rightarrow x=-0.5,0,2
second derivative= 12x^2 -12x -4
putting values of x in second derivative:

-0.5 \rightarrow 5
0 \rightarrow -4
2 \rightarrow 20
 hence there is a min point at -0.5 and 2 and max at 0, putting these values of x gives values of min points (-0.5, 7.8) and (2, 0), therefore as none of y values are below x axis, f(x) is always positive!  :D

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Offline SGVaibhav

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Re: Pure math(p3)
« Reply #34 on: May 28, 2009, 05:41:39 pm »
this is an A level thread, hmm, i was thinking that why dont i even know the head and tail of this.
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Offline astarmathsandphysics

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Re: Pure math(p3)
« Reply #35 on: May 28, 2009, 05:43:36 pm »
Quote from: ITS MEE... on May 28, 2009, 02:43:25 pm
can u integrate
sin^3xcosx
please :'(

orry i missed this earlier sub u=sinx then du=cosxdx so dx=1/cosxdu so you integrate

sin^3xcosxdx=sin^3xcosx(1/cosxdu)=u^3du and the ans =1/4u^4=1/4sin^4x
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Offline louis

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Re: Pure math(p3)
« Reply #36 on: May 28, 2009, 05:52:53 pm »
ITS MEE,

Attachment  (  complex  number )  for your attention.
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Offline twilight

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Re: Pure math(p3)
« Reply #37 on: May 28, 2009, 05:59:46 pm »
Quote from: astarmathsandphysics on May 28, 2009, 05:43:36 pm
Quote from: ITS MEE... on May 28, 2009, 02:43:25 pm
can u integrate
sin^3xcosx
please :'(

orry i missed this earlier sub u=sinx then du=cosxdx so dx=1/cosxdu so you integrate

sin^3xcosxdx=sin^3xcosx(1/cosxdu)=u^3du and the ans =1/4u^4=1/4sin^4x

thksssss astar .. +rep
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Offline louis

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Re: Pure math(p3)
« Reply #38 on: May 28, 2009, 06:05:01 pm »
ITS mee,


Another attachment  (  vertices of triangle ).
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Offline LupeFiasco

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Re: Pure math(p3)
« Reply #39 on: May 28, 2009, 11:33:30 pm »
anybody here taking bio P4, please send me what's coming for the exam at genetick2004@hotmail.com . Thanks and god speed.
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Offline louis

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Re: Pure math(p3)
« Reply #40 on: May 29, 2009, 07:29:53 am »
crucio,
I like your method by calculus calculation.
I make the graph based on your calculation.
Please look at the attachment.
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