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correct ordering of 1/x,2x and x^2

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correct ordering of 1/x,2x and x^2

by jain2016 » Wed May 04, 2016 10:41 pm
If x is positive, which of the following could be the correct ordering of 1/x,2x and x^2

1) x^2<2x<1/x

2) x^2<1/x<2x

3) 2x<x^2<1/x

A) None

B) 1 only

C) 3 only

D) 1 and 2 only

E) 1, 2, and 3

OAD

Hi Experts ,

Please explain.

Thanks,

SJ
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by Brent@GMATPrepNow » Thu May 05, 2016 9:32 am
If x is positive, which of the following could be the correct ordering of 1/x, 2x and x²?
I. x² < 2x < 1/x
II. x² < 1/x < 2x
III. 2x < x² < 1/x

(A) None
(B) I only
(C) III only
(D) I and II only
(E) I II and III
Let's start by plugging in some positive values of x and see what we get.

x = 1/2
1/x = 2
2x = 1
x² = 1/4
So, we get x² < 2x < 1/x
This matches statement I.

x = 3/4
1/x = 4/3
2x = 3/2
x² = 9/16
So, we get x² < 1/x < 2x
This matches statement II

x = 3
1/x = 1/3
2x = 6
x² = 9
So, we get 1/x < 2x < x²
NO MATCHES

At this point, the correct answer is either D or E.
If you're pressed for time, you might have to guess.

Alternatively, you can use some algebra to examine statement III (2x < x² < 1/x)
Notice that there are 2 inequalities here (2x < x² and x² < 1/x)
Take 2x < x² and divide both sides by x to get 2 < x
Take x² < 1/x and multiply both sides by x to get x^3 < 1, which means x < 1
Hmmm, so x is greater than 2 AND less than 1. This is IMPOSSIBLE, so statement III cannot be true.

Answer = D

Cheers,
Brent
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by 800_or_bust » Thu May 05, 2016 10:36 am
jain2016 wrote:If x is positive, which of the following could be the correct ordering of 1/x,2x and x^2

1) x^2<2x<1/x

2) x^2<1/x<2x

3) 2x<x^2<1/x

A) None

B) 1 only

C) 3 only

D) 1 and 2 only

E) 1, 2, and 3

OAD

Hi Experts ,

Please explain.

Thanks,

SJ
I came up with (D).

(I) and (II) can easily be proven by testing values. If x = 1/2, then x^2 = 1/4, 2x = 1, and 1/x = 2. So (I) could be true. If x = 3/4, then x^2 = 9/16, 1/x = 1.333, and 2x = 1.5, so (II) could be true.

(III) can never be true. In order for x^2 to be less than 1/x, we know we need to be dealing with a fractional value for x. However, if this is true x^2 is essentially one fraction times another fraction which will yield an even smaller fraction than x. 2x on the other hand is multiplying a fraction by the integer 2, which will result in a larger fraction than x (or perhaps even a mixed number). Therefore, for 0<x<1, 2x cannot be less than x^2.
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by jain2016 » Thu May 05, 2016 10:10 pm
Let's start by plugging in some positive values of x and see what we get.

x = 1/2
1/x = 2
2x = 1
x² = 1/4
So, we get x² < 2x < 1/x
This matches statement I.

Hi Brent ,

Thanks for your reply. Just a quick question x=1/2 doesn't matches the statement 2, then how come statement 2 matches?

Please explain and correct me if took in another way.

Many thanks in advance.

SJ
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by nchaswal » Fri May 06, 2016 12:56 pm
Hi there

The question asks which COULD be true not necessarily it has to be true always. BRENT took some pair of values and showed you which statements CAN be true(not necessarily for every number).
Statement III can never be true as if 1/x is greatest of them all then x has to be less than 1 and for this x^2 can never be greater than 2x as 2x is double of that number where as x^2 you are taking FRACTION of a FRACTION.
Hence only Statement I and II COULD be possible.

Statement 2 will match for any value where 1/sqrt(2)<x<1 . You can try this yourself by taking 9/10 or 0.9 as x
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