Exponents

moneyman · Posted: Tue May 06, 2008 5:37 am Post subject: Exponents

If x is positive, which of the following could be the correct ordering of 1/x,2x and x^2 ?

I.x^2<2x<1/x

II.x^2<1/x<2x

III.2x<x^2<1/x

None
I only
III only
I and II only
I II and III

Ans D

I picked a fraction and proved I right but as I started picking values I realised I am using more than 2 mins for this question..Any easier way to solve this problem??

VP_Jim · Posted: Tue May 06, 2008 8:07 pm Post subject:

Hi moneyman,

First off, picking fractions or any number, really (as long as it's positive) is the right way to approach this problem, so great job!

A quicker way to look at this is that you don't really need to do the calculations and solve everything. If you practice just looking at the problems and figuring out how to plug in numbers and do some estimating in your head (without actually having to write them out and solving for the numbers), you can save a lot of time.

Hope this helps!

akshatsingh · Posted: Wed May 07, 2008 12:26 am Post subject:

Just a quick question,

II.x^2<1/x<2x

how is this true if x=1/2 ?

pls suggest some value for which this is true..

Thanks

moneyman · Posted: Wed May 07, 2008 6:14 am Post subject:

Ya akshat I had the same doubt..need help with the second choice guys!!

dk_gmat · Posted: Wed May 07, 2008 9:15 am Post subject:

Ak/Maxx,

The relation x^2<1/x will only work for x <1> 1 this is not going to work.

As per your assumption of x = 1/2

If you deduce the value as
x^2 = 1/4 = 0.25
1/x = 2

and 0.25< 2

Hope this helps.

DK ... Very Happy

moneyman · Posted: Wed May 07, 2008 9:33 am Post subject:

Well dk you have missed out on 2x and have only calculated for 1/x and x^2..If x=1/2 then 2x=1 which is not greater than 1/x(according to stat 2)

VP_Jim · Posted: Wed May 07, 2008 3:08 pm Post subject:

The number 0.9 works with Statement II. It took me a few tries to get that Smile

netigen · Posted: Wed May 07, 2008 5:27 pm Post subject:

Figuring out why eq 2 holds is tricky.

Important point to note is that we have to prove 1/x<2x for a fraction between 0 and 1

as we already know that x^2<2x and x^2<1/x holds for all fractions between 0 and 1 (from eq 1)

1/x <2x or x^2 > 1/2, which equates to x>1/sqrt(2) ~ 0.71

so all numbers between 0.71 and 1 will satisfy this condition.