If x is positive, which of the following

[VJesus12]

If x is positive, which of the following could be correct ordering of $$\frac{1}{x},\ 2x\ ,\ \ \ \ and\ \ x^2\ ?$$ $$(I)\ \ \ \ x^2<2x<\frac{1}{x}$$ $$(II)\ \ \ x^2<\frac{1}{x}<2x$$ (III) 2x <x^2 < 1/x.

(A) none
(B) I only
(C) III only
(D) I and II
(E) I, II and III

The OA is the option D.

How can I know the correct answer? Experts, I need your help here.

[EconomistGMATTutor]

Hello Vjesus12.

First of all, I want to comment that the third option is not displayed. The third option is: $$(III)\ \ \ 2x<x^2<\frac{1}{x}$$

Let's try with x=1/2. Then $$\frac{1}{x}=\frac{1}{\frac{1}{2}}=2\ \ \ \ \ \ \ \ \ 2x=2\cdot\frac{1}{2}=1\ \ \ \ \ \ \ \ \ \ x^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}.$$ Therefore, the option (I) is possible.

If x=8/9 then $$\frac{1}{x}=\frac{1}{\frac{8}{9}}=\frac{9}{8}=1.125\ \ \ \ \ \ \ \ \ 2x=2\cdot\frac{8}{9}=\frac{16}{9}=1.777\ \ \ \ \ \ \ \ \ \ x^2=\left(\frac{8}{9}\right)^2=\frac{64}{81}=0.79.$$ Therefore, the option can be possible too.

When x is between 1 and 2, the order is 1/x < x^2 < 2x and when x is greater than 2, the order is 1/x < 2x < x^2, but these two options are no in the list.

So, the option (III) can not be possible.

So, the correct answer is [spoiler]D=(I) and (II)[/spoiler].

I hope this explanation may help you.

Feel free to ask me again if you have any doubt.

Regards.

[Scott@TargetTestPrep]

If x is positive, which of the following could be correct ordering of $$\frac{1}{x},\ 2x\ ,\ \ \ \ and\ \ x^2\ ?$$ $$(I)\ \ \ \ x^2<2x<\frac{1}{x}$$ $$(II)\ \ \ x^2<\frac{1}{x}<2x$$ (III) 2x <x^2 < 1/x.

(A) none
(B) I only
(C) III only
(D) I and II
(E) I, II and III

The OA is the option D.

How can I know the correct answer? Experts, I need your help here.

First, if x = 1/2, then 1/x = 2, 2x = 1, and x^2 = 1/4; i.e. x^2 < 2x < 1/x. Thus, Roman numeral I is possible. We eliminate answer choice A.

Looking at the answer choices, we see that 1/x is greater than x^2 in every choice; therefore it must be true that x < 1.

For Roman numeral II, in order for 1/x < 2x to hold, we must have 2x^2 > 1, which is equivalent to x^2 > 1/2. This implies that x > 1/√2 or x < -1/√2 (which is not possible because x is positive). Since √2 is roughly 1.41, we see that if we let x = 1/1.4 = 5/7, then we have 1/x = 7/5, 2x = 10/7, and x^2 = 25/49. In this case, we have x^2 < 1/x < 2x and thus, Roman numeral II is also possible.

Finally, for Roman numeral III to be true, we must have x^2 > 2x or equivalently, x^2 - 2x > 0. Factoring the left hand side, we get x(x - 2) > 0. In order for the product of x and (x - 2) to be positive, either both of them must be positive or both of them must be negative. If both x and (x - 2) are positive, then x > 2; but then x^2 > 1/x. If both of them are negative, then x < 0 but this contradicts the fact that x is positive. Therefore, Roman numeral III is not possible.

Answer: D