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Is x^2 > x? (1) x^2 > 1 (2) x > -1

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Joined: 02 Sep 2009
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Is x^2 > x? (1) x^2 > 1 (2) x > -1  [#permalink]

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New post 04 Dec 2017, 00:04
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

64% (01:02) correct 36% (00:58) wrong based on 42 sessions

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Re: Is x^2 > x? (1) x^2 > 1 (2) x > -1  [#permalink]

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New post 04 Dec 2017, 00:33
statement 1 is suff.
the only interval where x^2<x is when x is between 0 and 1.
if x^2>1, this translates to the fact that x is not between 0 and 1.

hence, A.
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Is x^2 > x? (1) x^2 > 1 (2) x > -1  [#permalink]

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New post Updated on: 04 Dec 2017, 04:21
Is x^2 > x?

Rephrase question : x(x-1) >0 ?
(1)\(x^2\) > 1
(x - 1)(x + 1)>0
range: (-infinity ,-1) U (1, infinity)
Substitute values from this range into the question and check whether we get positive value
x= -2
(-2)*(-3) >0 ? ..........yes
x=1.5
(1.5*0.5) >0 ? ...........yes
Sufficient

(2) x > -1
substitute 0.5
(.5)(-.5) >0? ....no
substitute -0.5
(-0.5)(-1.5) >0? ...yes

two different answers--> not sufficient

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Originally posted by TaN1213 on 04 Dec 2017, 02:15.
Last edited by TaN1213 on 04 Dec 2017, 04:21, edited 1 time in total.
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Re: Is x^2 > x? (1) x^2 > 1 (2) x > -1  [#permalink]

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New post 04 Dec 2017, 04:05
Question ask if the range of x

X<0 and x>1

St1:
X>1
X<-1

Sufficient

St2:
X>-1
X=0 (NO)
X=4 (YES)

Not sufficient

Therefore answer is A

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Re: Is x^2 > x? (1) x^2 > 1 (2) x > -1  [#permalink]

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New post 04 Dec 2017, 04:24
x^2 > x
or x(x-1)>0
or the question is asking is x>1 or x<0
Statement 1
x^2>1
(x-1)(x+1)>0
either x>1 or x<-1 in both the case it is sufficient
statement 2
x>-1
not sufficient , since x can be 0 (which doesn't satisfy my equation )
Hence IMO A
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Re: Is x^2 > x? (1) x^2 > 1 (2) x > -1 &nbs [#permalink] 04 Dec 2017, 04:24
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