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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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.
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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 ?
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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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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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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

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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The reasoning is that in these ranges x (2x), 1/x and x^2 are ordered differently:

For \(x>2\) --> \(x^2\) has the largest value. Since no option offers this we know that \(x\) cannot be more that 2;
For \(1<x<2\) --> \(2x\) has the largest value, then comes \(x^2\). Since no option offers this we know that \(x\) cannot be from this range either;

So, we are left with last range: \(0<x<1\). In this case \(x^2\) has the least value. Options, I and II offer this, so we should concentrate on them and test the values of x from 0 to 1.

Hope it's clear.
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Bunuel wrote:
Basically we should determine relationship between , and in three areas: 0------1------2-------.

Buneul,

Could you please explain as to how you came to pick these ranges ? The rest of it is perfectly fine.

Thanks
d
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Hello Bunuel,
Can you show us graphical approach to this question.
I was able to draw the graph for all three equations and intersection points, however, I was not able to negate the third ordering. Would you please help me out.

Thanks
imhimanshu
P.S - How can I post graphs here.