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![[Post New]](/templates/default/images/icon_minipost_new.gif) 1 Dec 2007 20:32:03 IST
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the solution of 2x+2IxI 2 2
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1 Dec 2007 22:41:30 IST
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I got [ log( 2+1)/log2 , infinity) is it correct...............?
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my future signature:
LOKESH SARDANA,
department of Production and Industrial engineering,
Indian Institute of Technology,Roorkee.
Happiness can be found, even in the darkest of times, if one only remembers to turn on the light.
Albus Dumbledore
Harry Potter and the Prisoner of Azkaban movie
There are only two ways to live your life. One is as though nothing is a miracle. The other is as though everything is a miracle.
-- Albert Einstein
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8 Dec 2007 20:42:38 IST
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no it is (- ,log2( 2 -1)] U[1/2, infinity)
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its simpl yaar in dis problem 2 cases arises 1 is x<0 n x>0
1stly v take x>0 in dis v get 2*2^x>=2rt2 hence x>=1/2 der4
xbelongs 2 [1/2, infinity)
next x<0 , 2^x +2^(-x)>=2rt2
let 2^x=y hence ,
y^2-2rt2 y +1>=0
hence by dis v get 2 values of y ie
y>= -1+rt2 n 1+rt2
hence by comparig d values v get x belongs 2
(infinity,log(-1+rt2)] U [1/2,infinity) rate me if u like it
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20 Dec 2007 11:33:59 IST
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THE BEST METHOD IS SOLVE USING ..........GRAPH FORM TO DIFFERENT EQUATION USING x>0 and x<0 . SKETCH THE CURVES AND GET THE ANS ..................................
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