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inequality

by sunilrawat » Fri Jul 01, 2011 9:41 pm
I have a doubt.
what could be the possible value if 'a' if
(a^3-a)<0

whats wrong with this ..?
a(a^2-1)<0
a(a+1)(a-1)<0
thus, a<0, a<-1, or a<1
combining, we get a<0.
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by Frankenstein » Fri Jul 01, 2011 10:22 pm
sunilrawat wrote:I have a doubt.
what could be the possible value if 'a' if
(a^3-a)<0

whats wrong with this ..?
a(a^2-1)<0
a(a+1)(a-1)<0
thus, a<0, a<-1, or a<1
combining, we get a<0.
Hi,
This is incorrect.
If (a+1)a(a-1) < 0
then a<-1, or 0<a<1
If you want to check, Consider any number between -1 and 0, say -1/2, the value of a(a^2-1) = (-1/2)(1/4-1) = 3/8>0
Cheers!

Things are not what they appear to be... nor are they otherwise
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by sunilrawat » Fri Jul 01, 2011 10:32 pm
This is exactly what I am not getting.
why didn't u consider a separately?
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by Frankenstein » Fri Jul 01, 2011 10:42 pm
sunilrawat wrote:This is exactly what I am not getting.
why didn't u consider a separately?
I don't understand what you mean by this statement. But, looking at your method, i think you have considered a<0 as a case.
I will solve this with your approach
case-1: a<0 and (a^2-1)>0
We get a<0 and (a<-1 or a>1)
So, the intersections part is a<-1, not a<0
case-2: a>0 and (a^-1)<0
we get a>0 and (-1<a<1)
So, the intersection part is 0<a<1

So, you get a<-1 or 0<a<1
Cheers!

Things are not what they appear to be... nor are they otherwise
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by sunilrawat » Fri Jul 01, 2011 10:58 pm
I got my mistake.
I didn't change inequality sign while shifting "-1" to the other side.

Thanks a lot.

Btw, it was a DS Q.
Is a<0?
a. (a^3-a)<0
b. (1-a^2)>0
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