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# Is x < 0 ?

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Manager
Joined: 10 Feb 2011
Posts: 114
Is x < 0 ? [#permalink]

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01 Mar 2011, 10:51
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Difficulty:

55% (hard)

Question Stats:

50% (01:57) correct 50% (00:52) wrong based on 68 sessions

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Is x < 0 ?

(1) x^3 < x^2
(2) x^3 < x^4
[Reveal] Spoiler: OA
Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 900
Re: Is x < 0 ? (1) x3 < x2 (2) x3 < x4 [#permalink]

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01 Mar 2011, 10:58
banksy wrote:
Is x < 0 ?
(1) x^3 < x^2
(2) x^3 < x^4

1) Insufficient. x < 0 or 0< x < 1
2) Insufficient. x < 0 or x > 1

1) + 2) Sufficient.

if 0 < x < 1 then x^4 < x^3 < x^2
if x < 0 then x^3 is negative and x^2 and x^4 are positive.

C
Manager
Joined: 14 Dec 2010
Posts: 206
Location: India
Concentration: Technology, Entrepreneurship
GMAT 1: 680 Q44 V39
Re: Is x < 0 ? (1) x3 < x2 (2) x3 < x4 [#permalink]

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02 Mar 2011, 00:20
I agree. C is correct.
Statement 1 tells us that x could be less than 0 or between 0 and 1. So insufficient.
Statement 2 tells us that x could be less than 0 or greater than 1. So insufficient.

Combining 1 and 2, we can deduce that x should be less than 0. So sufficient.
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Joined: 20 Dec 2010
Posts: 2010
Re: Is x < 0 ? (1) x3 < x2 (2) x3 < x4 [#permalink]

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02 Mar 2011, 01:04
banksy wrote:
Is x < 0 ?
(1) x^3 < x^2
(2) x^3 < x^4

Can someone explain how to find the ranges using the number line approach (+ve and -ve curves); I find these types so tough.

1.
x^3 < x^2
x^3 - x^2 < 0
x^2(x-1) < 0

Roots are; 0 and 1
Now; for a very small value of x; the above equation holds true

So; the ranges are x<0, 0<x<1, x>1

According to the rule; we should consider the -ve curves for "<" sign

so; x<0 and x>1. However, the correct ranges are x<0 or 0<x<1.

How come the positive curve 0<x<1 holds true for "<" sign and x>1 doesn't; How does the rule change for even exponents and odd exponents.

(2) x^3 < x^4

x^3 - x^4 <0
x^3(1-x) < 0
x^3(x-1) > 0 {Multiplying by -ve}

roots 0 and 1

for a big value of x, the above equation holds good
so; the ranges are; x>1, 0<x<1, x<0

Considering +ve curves because the sign here is ">"
x>1
x<0

This looks okay.
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Posts: 39622
Re: Is x < 0 ? (1) x3 < x2 (2) x3 < x4 [#permalink]

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02 Mar 2011, 02:44
2
KUDOS
Expert's post
fluke wrote:
banksy wrote:
Is x < 0 ?
(1) x^3 < x^2
(2) x^3 < x^4

Can someone explain how to find the ranges using the number line approach (+ve and -ve curves); I find these types so tough.

1.
x^3 < x^2
x^3 - x^2 < 0
x^2(x-1) < 0

Roots are; 0 and 1
Now; for a very small value of x; the above equation holds true

So; the ranges are x<0, 0<x<1, x>1

According to the rule; we should consider the -ve curves for "<" sign

so; x<0 and x>1. However, the correct ranges are x<0 or 0<x<1.

How come the positive curve 0<x<1 holds true for "<" sign and x>1 doesn't; How does the rule change for even exponents and odd exponents.

(2) x^3 < x^4

x^3 - x^4 <0
x^3(1-x) < 0
x^3(x-1) > 0 {Multiplying by -ve}

roots 0 and 1

for a big value of x, the above equation holds good
so; the ranges are; x>1, 0<x<1, x<0

Considering +ve curves because the sign here is ">"
x>1
x<0

This looks okay.

You should simplify x^3 < x^2 first. Reduce by x^2 (x^3 < x^2 means that x is not equal to zero so x^2>0) --> x<1, excluding one point x=0.

So you can ignore the even power expressions in the approach you talk (just look whether zero is inclusive or not). For example, solve (x+3)(x+1)(x-1)(x+5)^2<0:

Reduce by (x+5)^2 and note that x doesn't equal to -5'
So we have (x+3)(x+1)(x-1)<0 and x doesn't equal to -5 --> roots -3, -1 and 1. Trying extreme value for x tells that in 4th range (when x>1) the whole expression is positive so it's negative in 3rd and 1st ranges (+-+-): -1<x<1 or x<-3 excluding x=-5 (you can wirte it also as x<-5, or -5<x<-3, or -1<x<1).

If it were: (x+3)(x+1)(x-1)(x+5)^2<=0 then the answer would be: -1<=x<=1 or x<=-3 (so x=-5, as well as the other roots, is included as the whole expression can equal to zero)

Hope its' clear.
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Re: Is x < 0 ? [#permalink]

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04 Jun 2015, 00:51
C
From (1) x<0 or 0<x<1
From (2) x<0 or x>1

From both .... x<0
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Re: Is x < 0 ?   [#permalink] 04 Jun 2015, 00:51
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