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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Mar 2008 11:12:38 IST
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Try out this one :
f(x) =  [log x2 - 1 (x)]
Find the domain of the function.
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27 Mar 2008 12:51:57 IST
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edit.. complete stupidity.. thx madness
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JEE and OLYMPIA INFINATUM
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27 Mar 2008 13:04:59 IST
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Actually that wont work bcos for x belonging to (1,sqrt(2)) x^2-1 will be less than 1 and x>1 so the log value will become negative
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27 Mar 2008 13:52:37 IST
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the quantity inside the square root must be positive
x^2 - 1<= x
that implies x belongs to [(1-rt5)/2 , (1+rt5)/2]
and x^2 - 1 must be >1
that implies x>rt2
so the domain of the given function is (rt2,(1+rt5)/2]
thanq 'twas a good question
EDIT: Sorry, I found my mistake!!!
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27 Mar 2008 15:44:01 IST
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that is again stupidity. Why doesnt x=2 and all all? Personally i believe that the condn x^2 - 1 <= x is false, it isnt necessary :) the only necessary thing is that the parity of log x and logx^2-1 shd be the same and hence the answer is x belongs to (sqrt(2), infinity) :)
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27 Mar 2008 15:49:46 IST
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a) x>0
b) x^2-1>0 ,x^2-1  1(not equal to1)
c) when x>1,x^2-1 >1
d) when x<1 x^2-1<1
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by the left!!!
forward march!!!
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27 Mar 2008 15:50:11 IST
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wid this v can find domain
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by the left!!!
forward march!!!
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27 Mar 2008 15:51:00 IST
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plz verify whether my answer is correct 
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27 Mar 2008 15:53:19 IST
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yeah u r rite
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by the left!!!
forward march!!!
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27 Mar 2008 19:30:34 IST
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Rohit is absolutely right. Hence, the following must hold true  and The first set of conditions gives Now if  , The second possibility does not occur as  . For the first condition to be satisfied, we must have  Hence the domain is  |
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