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Is x^4-x>0?

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Is x^4-x>0?  [#permalink]

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New post 21 Oct 2016, 10:03
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  55% (hard)

Question Stats:

57% (01:50) correct 43% (01:29) wrong based on 180 sessions

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Is \(x^{4}\) - x >0 ?

1) x<0
2) \(x^{4} > x^{2}\)

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Re: Is x^4-x>0?  [#permalink]

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New post 21 Oct 2016, 10:45
idontknowwhy94 wrote:
Is \(x^{4}\) - x >0 ?

1) x<0
2) \(x^{4} > x^{2}\)

Rephrasing
x(x^3-1)>0
----> x<0 or x>1??

1) x<0 ---->suff
2)x^2(x^2-1)>0
x^2(x+1)(x-1)>0
thus -1<x<0 or x>1------->suff

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Re: Is x^4-x>0?  [#permalink]

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New post 21 Oct 2016, 21:02
1
idontknowwhy94 wrote:
Is \(x^{4}\) - x >0 ?

1) x<0
2) \(x^{4} > x^{2}\)


Is \(x^{4}\) - x >0 --> Is \(x^4 > x\) ?

St1: x < 0 --> x^4 is positive and x is negative. --> x^4 > x. Sufficient

St2: \(x^{4} > x^{2}\) --> If x^4 > x^2 then x^4 > x. Sufficient

Answer: D
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Re: Is x^4-x>0?  [#permalink]

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New post 21 Oct 2016, 21:14
idontknowwhy94 wrote:
Is \(x^{4}\) - x >0 ?

1) x<0
2) \(x^{4} > x^{2}\)


Question: Is \(x^{4}\) - x >0 ?

Here I am solving this question in more of logical approach rather than using more of Maths

x^4 will be greater than x only if
i) x>1 or
ii) x<0 or


Statement 1: x<0
SUFFICIENT

Statement 2: \(x^{4} > x^{2}\)
which is true only if
i) x>1 or
ii) x<-1 hence
SUFFICIENT

Answer: option D
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Re: Is x^4-x>0?  [#permalink]

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New post 22 Oct 2016, 01:15
idontknowwhy94 wrote:
Is \(x^{4}\) - x >0 ?

1) x<0
2) \(x^{4} > x^{2}\)


from stem is x(x^3-1)>0 true if x>1 or x<0 ...

from 1

x<0 ... suff

from 2

x^2(x^2-1) >0 , thus /x/>1, i.e. x<-1( X<0) or x>1 ..... suff

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Re: Is x^4-x>0?  [#permalink]

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New post 15 Jan 2019, 09:24
are we allowed to solve the second statement as follows:

\(x^4>x^2\) \(\implies\) \(x^2>0\) \(\implies\) \(x>0\)?
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Re: Is x^4-x>0?  [#permalink]

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New post 15 Jan 2019, 09:43
idontknowwhy94 wrote:
Is \(x^{4}\) - x >0 ?

1) x<0
2) \(x^{4} > x^{2}\)


\(x^{4}\) - x > 0
\(x^{4}\) will always be positive and for that inequality to be !> 0 , x has to be 1

1) x<0
this will take -ive values, which will always satisfy the original question.

2) \(x^{4} > x^{2}
[m]x^{4} - x^{2}\) > 0
\(x^{2}(x^{2} - 1)\) > 0
\(x^{2}\) will always be > 0, which means \((x^{2}\) > 1, this will take + ive values > than 1 , which will always satisfy the original question.

Answer D
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Re: Is x^4-x>0?   [#permalink] 15 Jan 2019, 09:43
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