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# Is x > x^2?

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Director
Joined: 07 Jun 2004
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24 Jan 2011, 08:38
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Difficulty:

45% (medium)

Question Stats:

63% (01:08) correct 37% (01:01) wrong based on 89 sessions

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Is x > x^2?

(1) x > x^3
(2) x > x^4
[Reveal] Spoiler: OA

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Re: last one for today DS algebra [#permalink]

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24 Jan 2011, 08:46
1
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Expert's post
rxs0005 wrote:
is x > x^2

S1 x > x^3

S2 x > x^4

Is $$x > x^2$$ --> is $$x(x-1)<0$$? --> is $$0<x<1$$?

(1) x > x^3 --> $$x*(x^2-1)<0$$ --> $$(x+1)*x*(x-1)<0$$ --> $$0<x<1$$ or $$x<-1$$. Not sufficient.

(2) x > x^4 --> $$0<x<1$$ (only in this range x will be more than x^4). Sufficient.

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Re: last one for today DS algebra [#permalink]

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24 Jan 2011, 19:41
Hi,

can you explain how you came up with this?
#1 above? specially the last part.

I was solving the other way. x > x^3 can have negative as well as positive.
Where as x>x^4 can only happen if 0<x<1.

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Re: last one for today DS algebra [#permalink]

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24 Jan 2011, 21:42
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Expert's post
rxs0005 wrote:
is x > x^2

S1 x > x^3

S2 x > x^4

A couple of things about x, x^2 and x^3:

1.
$$x^2 > x$$ when x > 1 or x < 0.
This implies that $$x^2 < x$$ when 0 < x< 1.
At 0 and 1, $$x^2 = x$$
(The relation is same for every positive even power.)

2.
$$x^3 > x$$ when x > 1 or -1 < x < 0
This implies that $$x^3 < x$$ when x < -1 or 0<x<1
At -1, 0 and 1, $$x^3 = x$$
(The relation is same for every positive odd power.)

Given: S1 $$x > x^3$$ means either x < -1 or 0<x<1
$$x > x^2$$ only when 0 < x< 1 hence statement 1 alone is not sufficient.

S2 x > x^4. This is obviously sufficient since the relation is same for every positive even power.

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Re: last one for today DS algebra [#permalink]

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25 Jan 2011, 03:23
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Expert's post
brandy96 wrote:
Hi,

can you explain how you came up with this?
#1 above? specially the last part.

I was solving the other way. x > x^3 can have negative as well as positive.
Where as x>x^4 can only happen if 0<x<1.

Check this: x2-4x-94661.html#p731476 or here: inequalities-trick-91482.html
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Re: last one for today DS algebra [#permalink]

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25 Jan 2011, 08:43
Thanks, This is great!!!

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Re: last one for today DS algebra [#permalink]

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25 Jan 2011, 12:49
to put it in words...

Question is actually whether x is a positive fraction (Simple Eg: 0.1 so that 0.1 > 0.01). So we need to find out if x is positive & if x lies between 0 & 1.

1) would hold good even when x is a negative integer => Not Suff
2) if x is greater than the even power of x it can only mean that x is positive and lies between 0 & 1 => Suff

So B.

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Re: last one for today DS algebra [#permalink]

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17 Mar 2011, 22:53
so in order for this to be true, x must be a positive fraction.
1) x>x^3
This says that x could be a positive fraction OR a negative integer.
Plug values to check for yourself.
2) x>x^4
This says that x definitely has to be a positive fraction.
Hence B

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Re: last one for today DS algebra [#permalink]

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18 Mar 2011, 00:15
x > x^2 when x is a +ve fraction

(1) x > x^3

not enough as -2 > -8 (-2)^3, or x can be a +ve fraction too

(2) x > x^4 when x is a +ve fraction, so sufficient.

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Re: Is x > x^2? [#permalink]

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24 Dec 2015, 15:05
is 0<x<1? other cases would yield a totally different outcome.

statement 1 tells that x>x^3. what if x=-3? then for sure x>x^3, but x is not > than x^2. in this case - the answer is no.
we can have x=0.25, thus, x>x^3. and the answer to the question is x>x^2 is yes.
since we have 2 outcomes, statement 1 alone is not sufficient.

statement 2 tells that x>x^4.
now, in this case, we know for sure that x is positive, and 0<x<1. which is sufficient.

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Re: Is x > x^2? [#permalink]

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22 Sep 2017, 07:28
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Re: Is x > x^2?   [#permalink] 22 Sep 2017, 07:28
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