Inequalities Made Easy!

If you are wondering about the absurd title of this post, just take a look at the above post'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.

Answer (D)

Is this method simpler?

Here, let’s look at a question that involves inequalities and modulus and is best understood using the concept of sets. It is not a difficult question but it is still very tricky. You could easily get it right the first time around but if you get it wrong, it could take someone many trials before he/she is able to convince you of the right answer. Even after I write a whole post on it, I wouldn’t be surprised if I see “but I still don’t get it” in the comments below!

Anyway, enough of introduction! Let’s get to the question now.

Question: If \(\frac{x}{|x|} < x\), which of the following must be true about x?
(A) x > 1
(B) x > -1
(C) |x|< 1
(D) |x| = 1
(E) |x|^2 > 1

Solution:
First thing we do is tackle the mod. We know that |x| is just the absolute value of x.

So, \(\frac{x}{|x|}\) can take only 2 values: 1 or -1
If x is positive, \(\frac{x}{|x|} = 1\) e.g. if \(x = 4\), then \(\frac{4}{|4|} = 1\)
If x is negative, \(\frac{x}{|x|} = -1\) e.g. if \(x = -4\), then \(\frac{-4}{|-4|} = \frac{-4}{4} = -1\)
x cannot be 0 because we cannot have 0 in the denominator of an expression.

Now let’s work on the inequality.

\(x >\frac{x}{|x|}\) implies \(x > 1\) if x is positive or \(x > -1\) if x is negative.

Hence, for this inequality to hold, either \(x > 1\) (when x is positive) or \(-1 < x< 0\) (when x is negative)

x can take many values e.g. -1/3, -4/5, 2, 5, 10 etc.

Now think – which of the following MUST BE TRUE about every value that x can take?
(A) x > 1
or
(B) x > -1

I hope that you agree that x > 1 doesn’t hold for every possible value of x whereas x > -1 holds for every possible value of x. Mind you, every value greater than -1 need not be a possible value of x.

This concept might need some more work. Let me explain with another example.

Forget this question for a minute. Say instead you have this question:

Example 1: x > 2 and x < 7. What integral values can x take?
I guess most of you will come up with 3, 4, 5, 6. That’s correct. I can represent this on the number line.



The top arrow shows x > 2 and the bottom arrow shows x < 7. You see that the overlapping area includes 3, 4, 5 and 6. That is the region that satisfies both the inequalities.

Now consider this:

Example 2: x > 2 or x > 5. What integral values can x take?
Let’s draw that number line again.



So is the solution again the overlapping numbers i.e. all integers greater than 5? No. This question is different. x is greater than 2 OR greater than 5. This means that if x satisfies at least one of these conditions, it is included in your answer. Think of sets. AND means it should be in both the sets (i.e. the overlapping part) as was the case in example 1. OR means it should be in at least one of the sets. Hence, which values can x take? All integral values starting from 3 onwards i.e. 3, 4, 5, 6, 7, 8, 9 …

Now go back to this question. The solution is a one liner.
\(\frac{x}{|x|}\) is either 1 or -1.
So \(x > 1\) or \(x > -1\)
So which values can x take? All values included in at least one of the sets. Therefore, x > -1.

Note that the confusion lies only between the first two options. The other three options are rejected outright.
(C) |x|< 1 implies -1 < x < 1. Definitely doesn’t hold.
(D) |x| = 1 implies x = 1 or -1. Definitely doesn’t hold.
(E) |x|^2 > 1 implies either x < -1 or x > 1. Definitely doesn’t hold.

So what do you say? Are you convinced that the answer is (B)? This question is discussed HERE.

Hi Where can i find more resources on modulus, word problems, number properties (odd/even, factors, LCM etc.) similar to the one above shared on Inequalities? Bunuel

Thank you