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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

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

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 !!:)

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

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

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...