If x is positive, which of the following could be correct ordering of

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

I. \(x^2<2x<\frac{1}{x}\)

II. \(x^2<\frac{1}{x}<2x\)

III. \(2x<x^2<\frac{1}{x}\)

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

First note that we are asked "which of the following COULD be the correct ordering" not MUST be.
Basically we should determine relationship between \(x\), \(\frac{1}{x}\) and \(x^2\) in three areas: \(0<1<2<\).

\(x>2\)

\(1<x<2\)

\(0<x<1\)

When \(x>2\) --> \(x^2\) is the greatest and no option is offering this, so we know that x<2.
If \(1<x<2\) --> \(2x\) is greatest then comes \(x^2\) and no option is offering this.

So, we are left with \(0<x<1\):
In this case \(x^2\) is least value, so we are left with:

I. \(x^2<2x<\frac{1}{x}\) --> can \(2x<\frac{1}{x}\)? Can \(\frac{2x^2-1}{x}<0\), the expression \(2x^2-1\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers)

II. \(x^2<\frac{1}{x}<2x\) --> can \(\frac{1}{x}<2x\)? The same here \(\frac{2x^2-1}{x}>0\), the expression \(2x^2-1\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers)

Answer: D.

Second condition: \(x^2<\frac{1}{x}<2x\)

The question is which of the following COULD be the correct ordering not MUST be.

Put \(0.9\) --> \(x^2=0.81\), \(\frac{1}{x}=1.11\), \(2x=1.8\) --> \(0.81<1.11<1.8\). Hence this COULD be the correct ordering.

Hope it's clear.
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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. \(2x < x^2 < \frac{1}{x}\)

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


Algebraic approach is given in my solution. Here is number picking:

I. \(x^2<2x<\frac{1}{x}\) --> \(x=\frac{1}{2}\) --> \(x^2=\frac{1}{4}\), \(2x=1\), \(\frac{1}{x}=2\) --> \(\frac{1}{4}<1<2\). Hence this COULD be the correct ordering.

II. \(x^2<\frac{1}{x}<2x\) --> \(x=0.9\) --> \(x^2=0.81\), \(\frac{1}{x}=1.11\), \(2x=1.8\) --> \(0.81<1.11<1.8\). Hence this COULD be the correct ordering.

III. \(2x<x^2<\frac{1}{x}\) --> \(x^2\) to be more than \(2x\), \(x\) must be more than 2 (for positive \(x-es\)). But if \(x>2\), then \(\frac{1}{x}\) is the least value from these three and can not be more than \(2x\) and \(x^2\). So III can not be true.

Thus I and II could be correct ordering and III can not.

Answer: D.
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Let's look at this question logically. There will be some key takeaways here so don't focus on the question and the (long) solution. Focus on the logic.

First of all, we are just dealing with positives so life is simpler.
To compare two terms e.g. \(x^2\) and \(2x\), we should focus on the points where they are equal. \(x^2 = 2x\) holds when \(x = 2\).
When \(x < 2, x^2 < 2x\)
When \(x > 2, x^2 > 2x\)

Similarly \(1/x = x^2\) when \(x = 1\)
When \(x < 1, 1/x > x^2\).
When \(x > 1, 1/x < x^2\)

Going on, \(1/x = 2x\) when \(x = 1/\sqrt{2}\)
When \(x < 1/\sqrt{2}, 1/x > 2x\)
When \(x > 1/\sqrt{2}, 1/x < 2x\)

So now you know that:
If \(x < 1/\sqrt{2}\),
\(1/x > 2x, 1/x > x^2\) and \(x^2 < 2x\)
So \(x^2 < 2x < 1/x\) is possible.

If \(1/\sqrt{2} < x < 1\)
\(1/x < 2x, 1/x > x^2\)
So \(x^2 < 1/x < 2x\) is possible.

If \(1 < x < 2\)
\(1/x < 2x, 1/x < x^2, x^2 < 2x\)
So \(1/x < x^2 < 2x\) is possible.

If \(x > 2\)
\(1/x < 2x, 1/x < x^2, x^2 > 2x\)
So \(1/x < 2x < x^2\) is possible.

For no positive values of x is the third relation possible.


*Edited
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I would typically start here by seeing if I can spot any obvious x-values that satisfy any of the statements.
For statements both I and II, \(x^2\) is the smallest value.
This typically occurs when \(0<x<1\).
So let's test a value in that range.

If \(x = 0.1\), we get:
Statement I. \((0.1)^2 < 2(0.1) < \frac{1}{(0.1)}\), which simplifies to: \(0.01 < 0.2 < 10\) WORKS!!
So, statement I could be true, which means we can eliminate answer choices A and C, since they incorrectly state that statement I can't be true

Aside: Since \(x = 0.1\) makes statement I true, I know that it won't make the other two statements true since they have different orderings of \(x^2\), \(2x\) and \(\frac{1}{x}\)

Now let's analyze statement II: \(x^2 < \frac{1}{x} < 2x\)
Since we know \(x\) is positive, we can safely multiply all parts by \(x\) to get: \(x^3 < 1 < 2x^2\)
If \(x^3 < 1\), then we know \(x<1\).

What about this part of the inequality: \(1 < 2x^2\)
Divide both sides by \(2\) to get: \(\frac{1}{2} < x^2\)
Find the square root of both sides: \(\sqrt{\frac{1}{2}} < x\)
We can rewrite this as follows: \(\frac{1}{\sqrt{2}}<x\)

ASIDE: Before test day, be sure to memorize the following approximations:
√2 ≈ 1.4
√3 ≈ 1.7
√5 ≈ 2.2

Substitute 1.4 for √2 to get: \(\frac{1}{1.4}<x\)
Approximate: \(0.7 < x\)

When we combine our two conclusions we see that statement II is true when \(0.7 < x < 1\)

To confirm, let's plug \(x = 0.8\) into statement II to get: \((0.8)^2 < \frac{1}{0.8} < 2(0.8)\), which simplifies (approximately) to \(0.64 < 1.25 < 1.6\) WORKS!
So, statement II could be true, which means we can eliminate answer choice B, since it incorrectly states that statement II can't be true

Now onto statement III: \(2x < x^2 < \frac{1}{x}\)
Take this part of the inequality: \(2x < x^2\)
Subtract \(2x\) from both sides: \(0 < x^2 - 2x\)
Factor: \(0 < x(x- 2)\)
Since we already know \(x\) is positive, the only way \(x(x- 2)\) can be positive is if \(x > 2\)
However, if \(x > 2\), then there's no way that \(\frac{1}{x}\) can have the greatest value.
Therefore, statement III can never be true.

Answer: D
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If 0 < x < 1, for example, x = 1/2, then 1/x = 2, 2x = 1 and x^2 = 1/4. We see that x^2 < 2x < 1/x. Roman numeral I could be true. Also, if x = 3/4; then 1/x = 4/3, 2x = 3/2 and x^2 = 9/16. We see that x^2 < 1/x < 2x. Roman numeral II could also be true.

For Roman numeral III, observe that x^2 > 2x is equivalent to x^2 - 2x > 0 and we can factor out the x to obtain x(x - 2) > 0. Since x is positive, we can divide each side by x and we will get x - 2 > 0; in other words, x > 2.

On the other hand, if 1/x > 2x, then 1/x - 2x > 0; or, equivalently, (1 - 2x^2)/x > 0. Since x is positive, we can multiply each side by x to obtain 1 - 2x^2 > 0, which is equivalent to 2x^2 < 1. Then, x^2 < 1/2 and we know that this is only possible when x < 1.

Thus, the inequalities x^2 > 2x and 1/x > 2x cannot hold simultaneously for a positive x. Therefore, Roman numeral III is not possible.

In conclusion, I and II could be true.

Answer: D
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Solution:

Its a "Could be" true question =>if we can prove that a statement is true for one particular set of numbers
=> the statement could be true =>and hence is a correct answer.

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

If x =1/2 -> x^2 =1/4 ,1/x = 2 and 2x = 1

1/4 < 1< 2

=> x^2 <2x<1/x (I can be true)

=> A & C can be eliminated

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

=>x^2<1/x

=> x^3<1 (As we know the signs of both x & 1/x are positive)

=> x<1.

Also, 1/x<2x

=> 1<2(x^2)

=>x>1/sqrt(2).

=>Number that satisfies 1/sqrt(2)<x<1 will make the statement true and hence II could be true.

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

=> 2x <x^2

= > 2<x or x>2

If x>2 then 1/x has to be smaller than 2x and hence III is not a choice. (option d)

Devmitra Sen
GMAT SME

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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.

Since we eliminated the answer choice “none” and since 1/x is greater than x^2 in every Roman numeral; 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. The latter 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
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Copied from an other forum. Thought it might help someone. This is a great explanation...

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Each one of these gives you two inequalities. You know that x is positive, so you don't need to worry about the sign changing direction.

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

This means that x^2<2x so divide by x to get x<2. The second one tells you that 2x<1/x which simplifies to x < 1/sqrt(2). These can obviously both be satisfied at the same time, so (1) works.

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

This means that x^2<1/x which gives x^3<1, or x<1. The second half gives you 1/x<2x or 1<2(x^2) or x>1/sqrt(2). So any number that satisfies 1/sqrt(2)<x<1 will work.

(3) 2x<x^2<1/x. The first part gives 2x<x^2 or x>2. The second half gives x^2<1/x or x^3<1 or x<1. Since the regions x>2 and x<1 do not overlap, (3) can not be satisfied.


The Answer choice is (4), 1 and 2 only.

could be correct ordering

So if we can find any example that satisfy the inequation, that statement will be correct

(I) x = 0.1 => 0.01 < 0.2 < 10
(II) x= 1/2 => 1/4 < 1/2 < 1

(III)
2x < x^2 <=> x ( 2 -x) < 0, x > 0 then x > 2

with x > 2 ==> x^2 < 1/x <=> x^3 < 1 <=> x < 1

So (III) can't happen

The answer is D

Great explanation Bunuel, but for curiosity purpose, as I understand \(2x^2-1\) should be negative for this equation \(\frac{2x^2-1}{x}<0\) to be true. However if I algebraically find the values for which \(2x^2-1\) is negative, then on plugging those values in \(x^2<\frac{1}{x}<2x\) I do not find that the equation satisfies. Instead it is the value for which \(2x^2-1\) is positive, that I find the end quation satifies .... why is that ?

You are mixing I and II. If you find the values of \(x\) from the range \(0<x<1\) for which \(2x^2-1\) is negative then \(x^2<2x<\frac{1}{x}\) will hold true (not \(x^2<\frac{1}{x}<2x\)).

Below is number plugging method:

I. \(x^2<2x<\frac{1}{x}\) --> \(x=\frac{1}{2}\) --> \(x^2=\frac{1}{4}\), \(2x=1\), \(\frac{1}{x}=2\) --> \(\frac{1}{4}<1<2\). Hence this COULD be the correct ordering.

II. \(x^2<\frac{1}{x}<2x\) --> \(x=0.9\) --> \(x^2=0.81\), \(\frac{1}{x}=1.11\), \(2x=1.8\) --> \(0.81<1.11<1.8\). Hence this COULD be the correct ordering.

III. \(2x<x^2<\frac{1}{x}\) --> \(x^2\) to be more than \(2x\), \(x\) must be more than 2 (for positive \(x-es\)). But if \(x>2\), then \(\frac{1}{x}\) is the least value from these three and can not be more than \(2x\) and \(x^2\). So III can not be true.

Thus I and II could be correct ordering and III can not.

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

I. \(x^2 < 2x < \frac{1}{x}\)

II. \(x^2 < \frac{1}{x} < 2x\)

III. \(2x < x^2 < \frac{1}{x}\)

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

ALGEBRAIC APPROACH:

First note that we are asked "which of the following COULD be the correct ordering" not MUST be.
Basically we should determine relationship between \(x\), \(\frac{1}{x}\) and \(x^2\) in three areas: 0------1------2-------.

\(x>2\)

\(1<x<2\)

\(0<x<1\)

When \(x>2\) --> \(x^2\) is the greatest and no option is offering this, so we know that x<2.
If \(1<x<2\) --> \(2x\) is greatest then comes \(x^2\) and no option is offering this.

So, we are left with \(0<x<1\):
In this case \(x^2\) is least value, so we are left with:

I. \(x^2<2x<\frac{1}{x}\) --> can \(2x<\frac{1}{x}\)? Can \(\frac{2x^2-1}{x}<0\), the expression \(2x^2-1\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers)

II. \(x^2<\frac{1}{x}<2x\) --> can \(\frac{1}{x}<2x\)? The same here \(\frac{2x^2-1}{x}>0\), the expression \(2x^2-1\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers)

Answer: D.


NUMBER PLUGGING APPROACH:

I. \(x^2<2x<\frac{1}{x}\) --> \(x=\frac{1}{2}\) --> \(x^2=\frac{1}{4}\), \(2x=1\), \(\frac{1}{x}=2\) --> \(\frac{1}{4}<1<2\). Hence this COULD be the correct ordering.

II. \(x^2<\frac{1}{x}<2x\) --> \(x=0.9\) --> \(x^2=0.81\), \(\frac{1}{x}=1.11\), \(2x=1.8\) --> \(0.81<1.11<1.8\). Hence this COULD be the correct ordering.

III. \(2x<x^2<\frac{1}{x}\) --> \(x^2\) to be more than \(2x\), \(x\) must be more than 2 (for positive \(x-es\)). But if \(x>2\), then \(\frac{1}{x}\) is the least value from these three and can not be more than \(2x\) and \(x^2\). So III can not be true.

Thus I and II could be correct ordering and III can not.

Answer: D.

Hope it's clear.
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Hi Bunuel,

I have doubt !!!

Lets submit the values in the equations...lets take x= 3

then


I. x^2<2x<1/x ===> 9<6<1/3 which is not true
II. x^2<1/x<2x ===> 9<1/3<6 which is not true again
III. 2x<x^2<1/x ===> 6< 9> 1/3 which is true....

so, i believe only III is the ans

When picking a number, the most important thing is that we should try all possible numbers that could give us different answers. Let me explain by telling you how I would put in numbers and check.

I see \(\frac{1}{x}\), 2x and \(x^2\).
I know I have to try numbers from two ranges at least '0-1' and '>1' since numbers in these ranges behave differently.
Also, \(x^2\) is greater than 2x if x > 2 e.g. 3x3 > 2x3 but if x < 2, then \(x^2\) is less than 2x e.g 1.5 x 1.5 < 2 x 1.5. So, I need to try a number in the range 1 to 2 as well.
Also, 0 and 1 are special numbers, they give different results sometimes so I have to try those as well. Let's start:

0- Since x is positive, I don't need to try it.

1/2 - I get 2, 1 and 1/4. I get the order \(x^2\) < 2x < \(\frac{1}{x}\)

1 - I get 1, 2, 1 Now, what you need to notice here is that 2x > \(\frac{1}{x}\) whereas in our above result we got \(\frac{1}{x}\) > 2x. This means there must be some value between 0 and 1 where \(\frac{1}{x}\) = 2x. Anyway, that doesn't bother me but what I have to do now is take a number very close to 1 but still less than it.
I take 15/16 (random choice). I get 16/15, 15/8 and \((\frac{15}{16})^2\). The first two numbers are greater than 1 and the last one is less than 1. I get the order \(x^2 < \frac{1}{x} < 2x\)

Now I need to try 3/2. I get 2/3, 3 and 9/4. So the order is \(\frac{1}{x} < 2x < x^2\)

I try 3 - I get 1/3, 6, 9
For these numbers, \(x^2\) will be greatest but none of the options have it as the greatest term.

Only I. and II. match hence answer is (D)
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First of all we are asked "which of the following COULD be the correct ordering" not MUST be.

"MUST BE TRUE" questions:
These questions ask which of the following MUST be true, or which of the following is ALWAYS true no matter what set of numbers you choose. Generally for such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

As for "COULD BE TRUE" questions:
The questions asking which of the following COULD be true are different: if you can prove that a statement is true for one particular set of numbers, it will mean that this statement could be true and hence is a correct answer.

Also: how is III correct for x=3?

Hope it helps.
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Bunuel I have reasonably implemented the key values approach in all my inequalities problems but I couldn't understand how 0, 1, 2 can be inferred to be the keys in this problem.
Can you elaborate why you chose 0 1 2 ?

Regards,
Sameer

We should check which of the 3 statements COULD be the correct ordering.

Now, the same way as x and x^2 have different ordering in the ranges 0<x<1 and x>1, 2x and x^2 have different ordering in the ranges 1<x<2 (1/x<x^2<2x) and x>2 (1/x<2x<x^2). Next, you can see that no option is offering such ordering thus if there is correct ordering listed then it must be for the x-es from the range 0<x<1. So, if we want to proceed by number plugging we know from which range to pick numbers. Also as in this range x^2 is the least value we can quickly discard option III and concentrate on I and II.
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Hi Bunnel,

Could you please provide a reasoning to the below text... how did you find the range...Pls help

I am not sure how the OA is D