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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 May 2008 18:14:37 IST
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find the domain of  where a= x2-lxl-2
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28 May 2008 18:19:15 IST
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is the answer -2<=x<=2 ??
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domain wud be (-infinity,-2]U[2,infinity)
first assume let x>0
|x|=x
so for sqrt a to be defined thye given equatn shud be >0
x^2-x-2>0
x=(-infinty -1]U[2,infinity)
but since x>0 so solution wud be [2,infinity)
similarly whn x<0
eqn wud be x^2+x-2
this is >0 for x =(-infinity,-2]U[1,infinty)
but x<0
so solution is (-infinty,-2]
so domain wud be (-infinity,-2]U[2,infinity)
nudge me if u have any doubt
rate if useful
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28 May 2008 18:20:48 IST
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no thats what even im geting but the answer is R-(-2,2)
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28 May 2008 18:24:20 IST
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ohh we have to take2 cases
aditi is correct!!
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28 May 2008 18:26:22 IST
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R-(-2,2) means the same thing dear as i have given above
if u subtrat the set (-2,2) from the set of real numbers R u wud get the same thing ie.(-infinty,-2]U[2,infinity)
i hope its clear now
tell me if u still have doubt
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28 May 2008 18:28:21 IST
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ya ive seen it...i rated u also...
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28 May 2008 18:32:38 IST
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thanx..
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