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Is x > x^2?

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Is x > x^2? [#permalink]

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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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New post 24 Jan 2011, 08:46
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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.

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

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New post 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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New post 24 Jan 2011, 21:42
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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.

Answer (B).
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Re: last one for today DS algebra [#permalink]

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New post 25 Jan 2011, 03:23
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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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New post 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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New post 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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New post 17 Mar 2011, 22:53
Very simple question. Think about this:
Question asks is x>x^2?
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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New post 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.

Answer B.
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Re: Is x > x^2? [#permalink]

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New post 24 Dec 2015, 15:05
the main question asks:
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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Re: Is x > x^2?   [#permalink] 22 Sep 2017, 07:28
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