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Re: Is x positive? [#permalink]
29 Jul 2012, 00:02

4

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

jayoptimist wrote:

Is x positive?

(1) 1/(x+1) < 1 (2) (x-1) is a perfect square.

What is wrong with the following line of thought for condition 1

1/(x+1) < 1

1 < (x+1)

1 -1 < x+1-1

0 < x Hence x>0.

We can not multiple \(\frac{1}{x+1}< 1\) by \(x+1\) since we don't know whether this expression (\(x+1\)) is positive or negative, for the first case we should keep the same sign but for the second case we should flip the sign (when multiplying by negative number we should flip the sign of the inequity).

Is x positive?

(1) 1/(x+1) < 1. If \(x=10\) then the answer is YES, but if \(x=-10\) then the answer is NO. Not sufficient.

Or if you want to solve this inequality then: \(\frac{1}{x+1}< 1\) --> \(1-\frac{1}{x+1}>0\) --> \(\frac{x}{x+1}>0\) --> \(x>0\) or \(x<-1\), hence \(x\) could be positive as well as negative. Not sufficient.

(2) (x-1) is a perfect square. Given: \(x-1=\{perfect \ square\}\) --> \(x=\{perfect \ square\}+1\). Now, since perfect square is more than or equal to zero, then \(x=\{perfect \ square\}+1=non-negative+1=positive\). Sufficient.

Re: Is x positive? [#permalink]
10 Jan 2013, 01:34

Bunuel wrote:

jayoptimist wrote:

Is x positive?

(1) 1/(x+1) < 1 (2) (x-1) is a perfect square.

What is wrong with the following line of thought for condition 1

1/(x+1) < 1

1 < (x+1)

1 -1 < x+1-1

0 < x Hence x>0.

We can not multiple \(\frac{1}{x+1}< 1\) by \(x+1\) since we don't know whether this expression (\(x+1\)) is positive or negative, for the first case we should keep the same sign but for the second case we should flip the sign (when multiplying by negative number we should flip the sign of the inequity).

Is x positive?

(1) 1/(x+1) < 1. If \(x=10\) then the answer is YES, but if \(x=-10\) then the answer is NO. Not sufficient.

Or if you want to solve this inequality then: \(\frac{1}{x+1}< 1\) --> \(1-\frac{1}{x+1}>0\) --> \(\frac{x}{x+1}>0\) --> \(x>0\) or \(x<-1\), hence \(x\) could be positive as well as negative. Not sufficient.

(2) (x-1) is a perfect square. Given: \(x-1=\{perfect \ square\}\) --> \(x=\{perfect \ square\}+1\). Now, since perfect square is more than or equal to zero, then \(x=\{perfect \ square\}+1=non-negative+1=positive\). Sufficient.

Answer: B.

Hope it's clear.

Bunuel,

How did you arrive at x<-1 in the explanation below?

"\frac{x}{x+1}>0 --> x>0 or x<-1, hence x could be positive as well as negative. Not sufficient."

Re: Is x positive? [#permalink]
10 Jan 2013, 03:16

Expert's post

jgomey wrote:

Bunuel wrote:

jayoptimist wrote:

Is x positive?

(1) 1/(x+1) < 1 (2) (x-1) is a perfect square.

What is wrong with the following line of thought for condition 1

1/(x+1) < 1

1 < (x+1)

1 -1 < x+1-1

0 < x Hence x>0.

We can not multiple \(\frac{1}{x+1}< 1\) by \(x+1\) since we don't know whether this expression (\(x+1\)) is positive or negative, for the first case we should keep the same sign but for the second case we should flip the sign (when multiplying by negative number we should flip the sign of the inequity).

Is x positive?

(1) 1/(x+1) < 1. If \(x=10\) then the answer is YES, but if \(x=-10\) then the answer is NO. Not sufficient.

Or if you want to solve this inequality then: \(\frac{1}{x+1}< 1\) --> \(1-\frac{1}{x+1}>0\) --> \(\frac{x}{x+1}>0\) --> \(x>0\) or \(x<-1\), hence \(x\) could be positive as well as negative. Not sufficient.

(2) (x-1) is a perfect square. Given: \(x-1=\{perfect \ square\}\) --> \(x=\{perfect \ square\}+1\). Now, since perfect square is more than or equal to zero, then \(x=\{perfect \ square\}+1=non-negative+1=positive\). Sufficient.

Answer: B.

Hope it's clear.

Bunuel,

How did you arrive at x<-1 in the explanation below?

"\frac{x}{x+1}>0 --> x>0 or x<-1, hence x could be positive as well as negative. Not sufficient."

We have \(\frac{x}{x+1}>0\).

The roots are -1, and 0 (equate the expressions to zero to get the roots and list them in ascending order), this gives us 3 ranges: \(x<-1\), \(-1<x<0\), and \(x>0\).

Now, test some extreme value: for example if \(x\) is very large number then all two terms (x and x+1) will be positive which gives the positive result for the whole expression, so when \(x>0\) the expression is positive. Now the trick: as in the 3rd range expression is positive then in 2rd it'll be negative and finally in 1st it'll be positive again: + - +. So, the ranges when the expression is positive are: \(x<-1\) (1st range) and \(x>0\) (3rd range).

Re: Is x positive? [#permalink]
11 Jan 2013, 17:33

The light bulb clicked!!!!

Got it so the condition is met for all numbers less than -1 or greater than 0, but any number between 0 and -1 (Fraction) will result in a negative number.

But even without thinking about the possibility of a fraction, Statement 1 is insufficient because x>0 or x<-1 right?

I know this is a silly, but I didn't realize I could find the roots by doing this:

Re: Is x positive? [#permalink]
07 Jun 2014, 23:04

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