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Re: Is |p|^2<|p| ? [#permalink]
18 Apr 2013, 14:52
1
This post received KUDOS
/p/(/p/-1)<0 we cannot divide the inequality with /p/ since we don't know if /p/=0. from statement 1 ,we have that -1<=p<=1 insuff from statement 2, p<>+-1. insuff
combined p lies in(-1 U +1). p=0.5 ,we have YES,p=0 we have NO. insuff IMO E
Hey, Good Q ..... I'm bit confused ... Correct me if I'm wrong ....
The Question asks....
Is |p|^2<|p| ? , one must always keep in mind that Is |p| is always positive, no matter the sign of P ..Okay & second thing that p^2 will always remain +ve. Now, |p|^2<|p| will only be possible if |p| must be a proper fraction okay. So, we have to find if |p| is a proper fraction or not. okay.
now statement 1 says .......... p^2 =<1 ...... this means that P can be equal to +1, -1 or p can be equal to any proper fraction as P^2 must always be less than 1 only if P is a proper fraction. so from this we have three values of P... 1, -1 & any proper fraction... however this is insufficient because of multiple possibilities.
Now statement 2 says ....... p^2-1 is not equal to Zero. that means that p is not equal to +,-1 , therefore it is clearly insufficient. .....
Now, 1+2 ........ as statement 1 says that p can be 1 or -1 or any proper fraction & statement 2 says that P is not equal to +,-1 , therefore we can conclude that p = Proper fraction & if P is a Proper fraction whether +ve Proper fraction or -ve Proper fraction, ..... |p|^2<|p| will always remain true. Hence , C.
Please correct me if I'm wrong. Pls...... _________________
If you don’t make mistakes, you’re not working hard. And Now that’s a Huge mistake.
Re: Is |p|^2<|p| ? [#permalink]
19 Apr 2013, 10:54
2
This post received KUDOS
Hi manishuol,
lets see if I can remove your doubts. The graph will help a lot, scroll down it's \(|p|^2-|p|<0\) we have to look where it is negative!
\(|p|^2<|p|\), remember that both |p| and |p|^2 don't have to be positive, they can also equal 0.
"So, we have to find if |p| is a proper fraction or not. okay." Not correct, we must find out if p is in the interval \(-1<p<1\) AND \(\neq{0}\).
"this means that P can be equal to +1, -1 or p can be equal to any proper fraction" AND 0 I would add, p ranges in the interval \(-1\leq{p}\leq{1}\) "however this is insufficient because of multiple possibilities." Correct
"Now statement 2 is clearly insufficient" Correct
Question: is p \(-1<p<1\) AND \(\neq{0}\)? Statement 1: \(-1\leq{p}\leq{1}\) Statement 2 \(p\neq{-,+1}\)
1+2 \(-1<p<1\) But no one says anything about \(p=0\)
The point is all here: p could still equal 0, and both statement don't give us info about this possibility.
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Re: Is |p|^2<|p| ? [#permalink]
19 Apr 2013, 11:06
Zarrolou wrote:
Hi manishuol,
lets see if I can remove your doubts. The graph will help a lot, scroll down it's \(|p|^2-|p|<0\) we have to look where it is negative!
\(|p|^2<|p|\), remember that both |p| and |p|^2 don't have to be positive, they can also equal 0.
"So, we have to find if |p| is a proper fraction or not. okay." Not correct, we must find out if p is in the interval \(-1<p<1\) AND \(\neq{0}\).
"this means that P can be equal to +1, -1 or p can be equal to any proper fraction" AND 0 I would add, p ranges in the interval \(-1\leq{p}\leq{1}\) "however this is insufficient because of multiple possibilities." Correct
"Now statement 2 is clearly insufficient" Correct
Question: is p \(-1<p<1\) AND \(\neq{0}\)? Statement 1: \(-1\leq{p}\leq{1}\) Statement 2 \(p\neq{-,+1}\)
1+2 \(-1<p<1\) But no one says anything about \(p=0\)
The point is all here: p could still equal 0, and both statement don't give us info about this possibility.
-----------------------------------
Yeah You're certainly right. I missed out 0 as one of the possibility. .......I really appreciate your quick help.Thanks !! Brother !! _________________
If you don’t make mistakes, you’re not working hard. And Now that’s a Huge mistake.
Re: Is |p|^2<|p| ? [#permalink]
25 Jun 2013, 15:59
Is |p|^2<|p| ?
1.p^2≤1
2.p^2-1≠0
Is |p|^2<|p|? ==> is -1<p<1
1.) insufficient as p could be 1 or -1 meaning p|^2 could equal |p| or p|^2 could be less than |p| Insufficient.
2.) p^2-1≠0 insufficient. p could = 1/2 which when squared = 1/4 which doesn't = 0 and p^2 would be less than P. On the other hand P could = 10 and 10^2-1 > 10.
1+2) we know that -1<p<1 and that p^2-1≠0 so the range of values to test is between -1 and 1. Fine. For ALL values of p except -1, 0 and 1 is |p|^2<|p|. However, while -1^2 - 1 and 1^2 - 1 = 0 (which rules out p being -1 or 1) p could be 0 In which case p^2 = p or it could be a fraction in which p^2 is less than p!
Re: Is |p|^2<|p| ? [#permalink]
24 Jul 2014, 04:16
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