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
c. III
d. I and II
e. I, II, and III

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 left and right 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 values to the left and right of each critical point.

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 A, B and C.

In III, the largest value listed is 1/x.
For 1/x to be the largest value, x would have to be a fraction.
But if x is a fraction, then it is not possible that 2x<x², since doubling a fraction INCREASES its value, while squaring a fraction results in a SMALLER value.
Since III is not possible, eliminate E.

The correct answer is D.

For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, the p is

A: Between 2 & 10
B: Between 10 & 20
C: Between 20 & 30
D: Between 30 & 40
E: Greater than 40

Since the difference between them is 1, h(100) and h(100)+1 are consecutive integers.
Consecutive integers are COPRIMES: they share no factors other than 1.

Let's examine why:

If x is a multiple of 2, the next largest multiple of 2 is x+2.
If x is a multiple of 3, the next largest multiple of 3 is x+3.

Using this logic, if we go from x to x+1, we get only to the next largest multiple of 1.
So 1 is the only factor common both to x and to x+1.
In other words, x and x+1 are COPRIMES.

Thus:
h(100) and h(100)+1 are COPRIMES. They share no factors other than 1.

h(100) = 2 * 4 * 6 *....* 94 * 96 * 98 * 100
Factoring out 2, we get:
h(100) = 2^50 (1 * 2 * 3 *... * 47 * 48 * 49 * 50)

Looking at the set of parentheses on the right, we can see that every prime number between 1 and 50 is a factor of h(100).
Since h(100) and h(100)+1 are coprimes, NONE of the prime numbers between 1 and 50 can be a factor of h(100)+1.

Thus, the smallest prime factor of h(100) + 1 must be greater than 50.

The correct answer is E.

hi Mitch,

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.

can you please tell when to use this critical points concept? I tired trial and error method but 1 and 2 failed when the value is less than 1 and hold good when value is greater than 1. 3rd one failed in both conditions.

GMATGuruNY and Anurag@Gurome, thank you for both of you guys' quick response! They are inspiring to me.

But for the second question, I still don't see how II could be right. It is also not demonstrated in either of your answer. I know we can eliminate the other options for the purpose of this exam, but anyone can explain to me or give me an example in what case could x^2 < 1/x <2x ?

Thanks!

=

senmaru wrote:GMATGuruNY and Anurag@Gurome, thank you for both of you guys' quick response! They are inspiring to me.

But for the second question, I still don't see how II could be right. It is also not demonstrated in either of your answer. I know we can eliminate the other options for the purpose of this exam, but anyone can explain to me or give me an example in what case could x^2 < 1/x <2x ?

Thanks!

Please reread my post above, in which I showed that if x=3/4, then x² < 1/x < 2x.

Ah got it. Thank you! I really like your method of using critical points!