If you are wondering about the absurd title of this post, just take a look at last week’s title. It will make much more sense thereafter. This post is a continuation of last week’s post where we discussed number plugging. Today, as per students’ request, we will look at the inequalities approach to the same official question. You will need to go through our inequalities post to understand the method we will use here.

Recall that, given a < b, (x – a)(x – b) < 0 gives us the range a < x < b and (x – a)(x – b) > 0 gives us the range x < a or x > b.

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

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

(II) x^2 < 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

Solution:  The question has three complex inequalities. We will take each in turn. Note that each inequality consists of two more inequalities. We will split the complex inequality into two simpler inequalities e.g. x^2 < 2x < 1/x gives us x^2 < 2x and 2x < 1/x. Next we will find the range of values of x which satisfy each of these two inequalities and we will see if the two ranges have an overlap i.e. whether there are any values of x which satisfy both these simpler inequalities. If there are, it means there are values of x which satisfy the entire complex inequality too. Things will become clearer once we start working on it so hold on.

Let’s look at each inequality in turn. We start with the first one:

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

We split it into two inequalities:

(i) x^2 < 2x

We can rewrite x^2 < 2x as x^2 – 2x < 0 or x(x – 2) < 0.

We know the range of x for such inequalities can be easily found using the curve on the number line. This will give us 0 < x < 2.

(ii) 2x < 1/x

It can be rewritten as x^2 – 1/2 < 0 (Note that since x must be positive, we can easily multiply both sides of the inequality with x)

This gives us the range -1/?2 < x < 1/?2 (which is 0 < x < 1/?2 since x must be positive).

Is there a region of overlap in these two ranges i.e. can both inequalities hold simultaneously for some values of x? Yes, they can hold for 0 < x < 1/?2. Hence, x^2 < 2x < 1/x will be true for the range 0 < x < 1/?2. So this could be the correct ordering. Let’s go on to the next complex inequality.

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

Again, let’s break up the inequality into two parts:

(i) x^2 < 1/x

x^1 < 1/x is rewritten as x^3 – 1 < 0 which gives us x < 1.

(ii) 1/x < 2x

1/x < 2x is rewritten as x^2 – 1/2 > 0 which gives us x < -1/?2 (not possible since x must be positive) or x > 1/?2

Can both x < 1 and x > 1/?2 hold simultaneously? Sure! For 1/?2 < x < 1, both inequalities will hold and hence x^2 < 1/x < 2x will be true. So this could be the correct ordering too.

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

The inequalities here are:

(i)  2x < x^2

2x < x^2 can be rewritten as x(x – 2) > 0 which gives us x < 0 (not possible) or x > 2.

(ii) x^2 < 1/x

x^2 < 1/x gives us x^3 – 1 < 0 i.e. x < 1

Can x be less than 1 and greater than 2 simultaneously? No. Therefore, 2x < x^2 < 1/x cannot be the correct ordering.

Is this method simpler?

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

### 14 Responses

1. Jesse says:

Both are equally confusing and something I would not be able to perform on test day.

• Karishma says:

Yes, it is a very tricky question. Usually, in such questions, my first thought is number plugging. The first post just discusses how to ensure you have checked out the right values. Go through it a couple of times – you will get the hang of it. The transition points concept can help you in quite a few questions.

2. Jamin says:

Hi Karishma,

In the beginning of the article it is mentioned that when (x – a)(x – b) > 0 the range of x is x > a or x 0 -> x>a
x-b > 0 -> x>b

case 2: when both factors are negative
x-a x<a
x-b x<b

Could please explain,using the number line method , what am I missing?

• Jamin says:

Understood. Thank you :)!!

3. Puneet says:

Can u please explain how to solve this 1/x < 2x using the number line method.
thank u

• Karishma says:

1/x < 2x
Given x is positive, we multiply both sides with x.

1 0
(?2x +1)*(?2x – 1) > 0
Plot -1/?2 and 1/?2 on the number line. The positive region will be the rightmost region i.e. x > 1/?2 and leftmost region x < -1/?2.

Note that we could multiply by x because x is given to be positive. If this is not necessary, you must do this:

1/x – 2x < 0
(1 – 2x^2)/x 0

(?2x +1)*(?2x – 1)/x > 0

So we have 3 points: -1/?2, 0, 1/?2
Positive regions will be x > 1/?2 and -1/?2 < x <0

• Puneet says:

thank u i understood it

• Puneet says:

In the last part:
one thing i didnt understand is why u put >0 instead of < 0 as is in the question. Is it typo?

i understand that the positive regions will be as u wrote.

thank u

• Karishma says:

Do you mean 2x 0.

You move 2x to the right in the original inequality:
0 0

• Vaishno says:

Karishma,

1/x 0

Where is “x” in the above inequality?

Shouldn’t it be (?2x +1)*(?2x – 1)/x > 0 as you mentioned in the above post?

Is there a typo or am I missing something? Thanks.

• Karishma says:

1/x < 2x
You multiply both sides by x to get (note that x is positive)
1 < [2(x)^2]
[2(x)^2] – 1 < 0
and so on…

4. Shivani says:

Hi..
I have gone few of your post(quant) , just have 1 question – Can these methods be applied to generic questions or remain specific to the type of questions you have posted.

Thx

• Karishma says:

The methods discussed are usually quite generic. It will take some understanding and practice to know where you should use a particular method.