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Re: if -1<x<o, which of following must be true? [#permalink]

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31 Jan 2013, 06:28

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rakesh20j wrote:

if -1<x<o, which of following must be true? 1) x^3<x^2 2) x^5<1-x 3) x^4<x^2

A. I only B. I and II C. I II and III D. I and III E. II and III

The question should read:

If -1 < x < 0, which of the following must be true?

I. x^3 < x^2 II. x^5 < 1 – x III. x^4 < x^2

A. I only B. I and II only C. II and III only D. I and III only E. I, II and III

I. x^3 < x^2 --> from -1 < x < 0 it follows that LHS<0<RHS, so this statement is true.

II. x^5 < 1 – x --> x(x^4+1) < 1 --> negative*positive < 0 < 1, so this statement is also true.

III. x^4 < x^2 --> reduce by x^2 (we can safely do that since from -1 < x < 0 is follows that x^2>0): x^2 < 1. Again, as -1 < x < 0, then x^2 must be less than 1. Hence, this statement is also true.

Re: if -1<x<o, which of following must be true? [#permalink]

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01 Jun 2013, 03:49

Bunuel wrote:

rakesh20j wrote:

if -1<x<o, which of following must be true? 1) x^3<x^2 2) x^5<1-x 3) x^4<x^2

A. I only B. I and II C. I II and III D. I and III E. II and III

The question should read:

If -1 < x < 0, which of the following must be true?

I. x^3 < x^2 II. x^5 < 1 – x III. x^4 < x^2

A. I only B. I and II only C. II and III only D. I and III only E. I, II and III

I. x^3 < x^2 --> from -1 < x < 0 it follows that LHS<0<RHS, so this statement is true.

II. x^5 < 1 – x --> x(x^4+1) < 1 --> negative*positive < 0 < 1, so this statement is also true.

III. x^4 < x^2 --> reduce by x^2 (we can safely do that since from -1 < x < 0 is follows that x^2>0): x^2 < 1. Again, as -1 < x < 0, then x^2 must be less than 1. Hence, this statement is also true.

Answer: E.

Hope it's clear.

Bunuel,

Did not understand the colored part.

As -1 < x < 0, then X will always be negative and X2 will always be positive, so how to derive that x<1. _________________

"Where are my Kudos" ............ Good Question = kudos

Re: if -1<x<o, which of following must be true? [#permalink]

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01 Jun 2013, 03:57

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This post received KUDOS

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targetbschool wrote:

Bunuel wrote:

rakesh20j wrote:

if -1<x<o, which of following must be true? 1) x^3<x^2 2) x^5<1-x 3) x^4<x^2

A. I only B. I and II C. I II and III D. I and III E. II and III

The question should read:

If -1 < x < 0, which of the following must be true?

I. x^3 < x^2 II. x^5 < 1 – x III. x^4 < x^2

A. I only B. I and II only C. II and III only D. I and III only E. I, II and III

I. x^3 < x^2 --> from -1 < x < 0 it follows that LHS<0<RHS, so this statement is true.

II. x^5 < 1 – x --> x(x^4+1) < 1 --> negative*positive < 0 < 1, so this statement is also true.

III. x^4 < x^2 --> reduce by x^2 (we can safely do that since from -1 < x < 0 is follows that x^2>0): x^2 < 1. Again, as -1 < x < 0, then x^2 must be less than 1. Hence, this statement is also true.

Answer: E.

Hope it's clear.

Bunuel,

Did not understand the colored part.

As -1 < x < 0, then X will always be negative and X2 will always be positive, so how to derive that x<1.

Consider this: \(x^4 < x^2\) holds true if \(-1<x<0\) or \(0<x<1\).

Therefore, since given that \(-1<x<0\), then \(x^4 < x^2\) must be true.

Re: If -1 < x < 0, which of the following must be true? [#permalink]

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25 Jun 2014, 05:01

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If -1 < x < 0, which of the following must be true? [#permalink]

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15 Sep 2015, 04:03

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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A. I only B. I and II only C. II and III only D. I and III only E. I, II and III

Make sure to format the question properly. When you select the tag "source-others please specify" , do mention at the end of the question.

As for the question look below.

You are told that \(-1<x<0\) ---> x is a negative fraction less -1.

Let us analyse the 3 options.

I. \(x^3 < x^2\)----> MUST BE TRUE as \(x^2\) of \('x'\) in this question >0 while \(x^3 <0\). Consider \(x=-0.5\), \(x^2=0.25 >0\) while \(x^3=-0.125 <0\)[/m].

II. \(x^5 < 1 – x\) ----> MUST BE TRUE. Odd power of a negative umber \(x\) (=\(x^5\))<0 while 1-x for a negative \(x\) >0. Consider \(x=-0.5\) ---> \(1-(-0.5)=1+0.5=1.5 >0\)

III. \(x^4 < x^2\) -----> MUST BE TRUE. For fractions -1<x<0 or 0>x>1, greater the power, smaller will be the number. Consider \(x=-0.5\), \(x^2=0.25\), \(x^4 = 0.0625\), thus\(x^4<x^2\)

Thus, I,II,III are all MUST BE TRUE ----> E is the correct answer.

A. I only B. I and II only C. II and III only D. I and III only E. I, II and III

Hi, this Qs tests the properties of numbers/ fractions between 0 and 1..

so lets first check the properties, the answer will come automatically..

1) any even power will always be greater than any odd power.. x^even>x or x^odd..

2) within even powers-- as we keep increasing power the value will keep becoming lesser.. x^4<x^2..

3) Opposite will be true of ODD powers although the numeric value will be same as in even powers but due to a -ive sign, opposite is true.. x^3>x

4) Roots-- even roots will be imaginary.. \sqrt{x} not real Odd roots are possible and will be lesser for lower 3rd root will be more than 5th root of x.. or 3rd root of x<x

so lets see three choices now-- I. \(x^3 < x^2\) TRUE as per point 1) above

II.\(x^5 < 1 – x\) x is negative so -x will be positive and 1-x will be positive too.. whereas x^5 is -ive.. hence TRUE

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