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Plug in question

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by DevB » Tue Jul 22, 2014 5:16 am
The below question is from GMAT Prep practice exam so anyone planning to give the exam may avoid looking below.

For others, Please help in answering the following:

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
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by abhasjha » Tue Jul 22, 2014 5:20 am
solution by mitch hunt

Determine the critical points by setting the expressions equal to each other:

1/x = 2x
2x² = 1
x² = 1/2
x = √(1/2) = 1/√2 ≈ 1/1.4 ≈ 10/14 ≈ 5/7.

1/x = x²
x^3 = 1
x = 1.

2x = x²
x=2
(We can divide by x because x>0.)

The critical points are x=5/7, x=1, x=2.
These critical points indicate where two of the expressions are equal.
Thus, to the right and left of each critical point, the value of one expression must be greater than the value of another.

To determine which answer choices are possible, plug in one value to the left and one value to the right of each critical point.

x < 5/7:
If x=1/2, then:
1/x = 2.
x² = 1/4.
2x = 1.
Since x² < 2x < 1/x, we know that I could be true.
Eliminate A and C.

5/7 < x < 1:
If x = 3/4, then:
1/x = 4/3.
x² = 9/16.
2x = 3/2.
Since x² < 1/x < 2x, we know that II could be true.
Eliminate B.

In III, the largest value listed is 1/x.
For 1/x to be the largest value, x would have to be a fraction.
Having tried a fraction on each side of the critical point of 5/7, we know that there is no way that III could be true.

The correct answer is D.

https://www.beatthegmat.com/need-explanation-t85192.html
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by Brent@GMATPrepNow » Tue Jul 22, 2014 5:41 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 [Aside: We can safely do this because we are told that x is positive]
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
Brent Hanneson - Creator of GMATPrepNow.com
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by DevB » Tue Jul 22, 2014 6:15 am
Thanks Brent..!! I could find the plug in for first equation but was not able to prove the second equation correct.

Thanks again.
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