Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Solve the Inequality [#permalink]
04 Aug 2009, 03:56

1

This post received KUDOS

apoorvasrivastva wrote:

crejoc wrote:

X > 1/X

Kindly help to solve ... Thanks in Advance.

i will try and solve...Ian please comment

x>1/x x-(1/x)>0 (x^2-1)/x > 0

so EITHER 1/x > 0 and x^2 - 1 > 0 OR 1/x<0 and x^2 - 1 < 0 [ since (x^2-1)/x > 0 ...either both are positive or both are negative]

Case 1:

1/x is greater than 0 then x > 0 and for x^2 - 1 > 0 ... x >1 or x>-1

so finally x>1

Cae 2:

1/x < 0 then x is less than 0 and for x^2 - 1 < 0 ..... x<-1 and x <1

so finally x < -1

PLEASE CORRECT MY APPROACH

apoorvasrivastva, your approach is good for case 1, however for case 2 it will be -1 < x < 0 as nplaneta posted earlier.

The reason being... (x^2 - 1) / x > 0 When x < 0 then: (x^2 - 1) < 0 [multiplying both sides by x (which is a -ve number, hence flipping the inequality)] x^2 < 1 now as we are considering x < 0 then, when we take square root of both side: sqrt(x^2) < sqrt(1) x < 1 [but x is a negative number so we need to multiply both sides by -1] -x > -1 x > -1 [left side became positive due to negative x and -1] and since the case is for x < 0 , therefore, the solution is -1 < x < 0

The solution for x > 1/x being: x>1 (when x>0) and -1 < x < 0 (when x<0)

For reference and review of inequality principles, see basic-mathematical-principles post by HongHu in this forum [sorry I can't post the direct link, as the forum isn't allowing me cos I am new member] This exact problem is explained in this post as well.

~Cheers

_________________

The three most significant times in your life are: 1. When you fall in love 2. The birth of your first child 3. When you prepare for your GMAT

Re: Solve the Inequality [#permalink]
04 Aug 2009, 04:25

scarish wrote:

For reference and review of inequality principles, see basic-mathematical-principles post by HongHu in this forum

Actually i got this problem from the very link you mentioned of HongHu's post, as it is not explained in detail there , i came up posting it in a separate thread...

scarish wrote:

case 2 it will be -1 < x < 0 as nplaneta posted earlier. When x < 0 then: (x^2 - 1) < 0 [multiplying both sides by x (which is a -ve number, hence flipping the inequality)] x^2 < 1 .......... x < 1 [but x is a negative number so we need to multiply both sides by -1]

I couldn't understand why you have to multiply x, you said to flip the inequality, but it is not flipped i think. Then why have to multiply by -1. sorry i don't get you. May be i am not comprehending it.. Could you explain clearly, the second case. i was also getting the same answer as apoorvasrivastva

Re: Solve the Inequality [#permalink]
04 Aug 2009, 05:19

scarish wrote:

Cae 2:

1/x < 0 then x is less than 0 and for x^2 - 1 < 0 ..... x<-1 and x <1

so finally x < -1

PLEASE CORRECT MY APPROACH

apoorvasrivastva, your approach is good for case 1, however for case 2 it will be -1 < x < 0 as nplaneta posted earlier.

The reason being... (x^2 - 1) / x > 0 When x < 0 then: (x^2 - 1) < 0 [multiplying both sides by x (which is a -ve number, hence flipping the inequality)] x^2 < 1 now as we are considering x < 0 then, when we take square root of both side: sqrt(x^2) < sqrt(1) x < 1 [but x is a negative number so we need to multiply both sides by -1] -x > -1 x > -1 [left side became positive due to negative x and -1] and since the case is for x < 0 , therefore, the solution is -1 < x < 0

The solution for x > 1/x being: x>1 (when x>0) and -1 < x < 0 (when x<0)

For reference and review of inequality principles, see basic-mathematical-principles post by HongHu in this forum [sorry I can't post the direct link, as the forum isn't allowing me cos I am new member] This exact problem is explained in this post as well.

~Cheers[/quote]

Thanks i realsed my mistake!!!Kudos given to u !!:)

Re: Solve the Inequality [#permalink]
04 Aug 2009, 16:09

1

This post received KUDOS

scarish wrote:

case 2 it will be -1 < x < 0 as nplaneta posted earlier. When x < 0 then: (x^2 - 1) < 0 [multiplying both sides by x (which is a -ve number, hence flipping the inequality)] x^2 < 1 .......... x < 1 [but x is a negative number so we need to multiply both sides by -1]

crejoc wrote:

I couldn't understand why you have to multiply x, you said to flip the inequality, but it is not flipped i think. Then why have to multiply by -1. sorry i don't get you. May be i am not comprehending it.. Could you explain clearly, the second case. i was also getting the same answer as apoorvasrivastva

Right, our equation was (x^2-1)/x > 0, so to get rid of x from the denominator you have to multiply both sides of the inequality by x, that is (x^2-1)*x/x > 0*x => (x^2-1)/x < 0 and the sign is flipped because x < 0. Multiplying both sides by x is what you do when we simply "cross-multiply" but we can only safely cross-multiply when it is equality but not in inequality.

Secondly, upon solving our equation we arrived at: x < 1

Now x in this equation is -ve, since we are considering a case where x < 0. So if we leave it like x < 1 then the x in this case is not comparable with x in our solution for the case where x > 0. So in order to compare apples with apples, we need to get x to mean the same thing for both cases. Therefore, we multiply both sides by -1 (which flips the sign again) to arrive at: x > -1

Also, try picking few numbers in and outside the boundaries in the original equation, which may also help.

crejoc wrote:

Actually i got this problem from the very link you mentioned of HongHu's post, as it is not explained in detail there , i came up posting it in a separate thread...

On a separate note, I think if you have doubts on questions in a particular post then its better to post your doubts on that post itself, so as to keep all responses on the issue together, instead of starting the new post. I don't know if this is the policy on this forum or not, may be moderator can clarify?

_________________

The three most significant times in your life are: 1. When you fall in love 2. The birth of your first child 3. When you prepare for your GMAT

Re: Solve the Inequality [#permalink]
07 Aug 2009, 01:51

i think best approach in those 2 mins in GMAT will be to plug values.....just keep in mind that whenever it's about fraction....intervals should be 1<x; 0<x<1; -1<x<0; -1>x

we find only applicable are 1<x and -1<x<0

i hope this helps

_________________

Bhushan S. If you like my post....Consider it for Kudos

Re: Solve the Inequality [#permalink]
09 Aug 2009, 04:24

bhushan252 wrote:

i think best approach in those 2 mins in GMAT will be to plug values.....just keep in mind that whenever it's about fraction....intervals should be 1<x; 0<x<1; -1<x<0; -1>x

we find only applicable are 1<x and -1<x<0

i hope this helps

Yes the faster the better, thanks for your approach...