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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 May 2008 20:37:55 IST
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however i have got the solutions , i cant get the thing behind it ....::
f(x)=(2x/2x^2+5x+2) ; g(x)=(1/1+x):
find the set of real values of x for which f(x)>g(x) ??
plzzz give the right concept behind it....>??
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15 May 2008 20:38:48 IST
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i cant see a bit of it........make it better dude
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15 May 2008 20:40:09 IST
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SET OF VALUES : FOR f(x)>g(x).....
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15 May 2008 20:49:45 IST
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2x(1+x)>2x^2+5x+2
Thus,x<-2/3 comes out to be the solution.
Thus,x  (- ,-2)U(-2,-1)U(-1,-2/3)
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MaNuTd RoXxXx..MaNuTd 2 WiN PrEmIeR LeAgUe ThIs SeAsOn ToO AlOnG WiTh ChAmPiOnS LeAgUe.....HaiL RoNaLdO ...HaiL LaMpArD... |
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15 May 2008 20:53:14 IST
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no yaar lampard ....the answer is partially correct...??
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15 May 2008 21:14:47 IST
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rates assured....!!!!
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15 May 2008 21:23:41 IST
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What is the answer then??I think this is the answer...
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MaNuTd RoXxXx..MaNuTd 2 WiN PrEmIeR LeAgUe ThIs SeAsOn ToO AlOnG WiTh ChAmPiOnS LeAgUe.....HaiL RoNaLdO ...HaiL LaMpArD... |
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15 May 2008 22:11:28 IST
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is the answer -x belongs to (-inf,-1)U(-1/2,1/2).
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15 May 2008 23:35:04 IST
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the answer is
x E (-2,-1)U(-2/3,-1/2)
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15 May 2008 23:37:16 IST
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but i want correct explanation....!!!!!!
i invite people to put on their comments
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15 May 2008 23:38:56 IST
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can u wait for a while ?
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SAIYANS ARE OF TRUE WARRIOR RACE . DONT UNDER-ESTIMATE US!
special theory of relativity ....i luv it...
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15 May 2008 23:41:31 IST
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ya sure..!!
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f(x) > g(x)
f(x) - g(x) > 0
2x/ (2x^2 + 5x+ 2) - 1 / (1+x) > 0
2x / (2x+1)(x+2) - 1/1+x > 0
-3x - 2 / (2x-1)(x+2)(x+1) > 0
3x+2 / (2x-1)(x+2)(x+1) < 0
x not equal to -1, -2, -1/2 as we get 0 in denominator... and x not equal to -2/3 as numerator becomes 0
Now apply wave curve method.
x E (-2, -1) U (-2/3, -1/2)
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* Gaurav Ragtah ( aka Artemis Fowl )
* Agent 'G' [sniper] - SD-6 (Alliance of Twelve)
* Your friendly neighborhood spideyunlimited |
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