If |x| < x^2, which of the following must be true ?

Can anyone please clarify my doubt here?

I generally solve such questions using number line after I have simplified to get individual values of x.

I fail to understand how x^2 has been eliminated here. What's troubling me here is that I denoted 'zero' on the number line.

Please help!

In the original question, we know |x| < x^2. So it's impossible that x = 0, because x = 0 doesn't satisfy that inequality.

There are a lot of ways to solve the original question, but if you did end up with the inequality

\(x^2(x^2 - 1)> 0\)

then since x^2 is always a positive number (we know x is not zero), we can safely divide both sides of the inequality by x^2. Since x^2 must be positive, we don't need to worry about whether we might be dividing by a negative number, so we don't need to worry about whether we should reverse the inequality. And once you divide by x^2, you learn x^2 - 1 > 0, or x^2 > 1.

I don't think that approach is the best one to take on questions like this, but I'll assume other approaches were already explained earlier in the thread (I haven't read all the solutions).
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Solution:

If x < 0, then -x is positive and that is why |x| is equal to -x. Let’s use some actual values to make it clear; for instance let x = -2. Now |-2| = 2 and 2 is equal to -x (because -x = -(-2) = 2).
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I find that these questions can be solved quickly (and accurately) by testing values.

For example, if |x| < x², then x could equal 2 (since |2| < 2²)
This means we can eliminate choices B and D since they state that x cannot equal 2.

Similarly, if |x| < x², then x could equal -2 (since |-2| < (-2)²)
This means we can eliminate choices A and C since they state that x cannot equal -2.

By the process of elimination, the correct answer is E

Given that |x| < x²

Let's take two cases to open the absolute value

Case 1: Whatever is inside the Absolute Value >= 0
x >= 0
=> |x| = x (Watch this video to learn more about Basics of Absolute Value)
=> x < \(x^2\)
=> \(x^2\) - x > 0
=> x*(x-1) > 0



=> x < 0 or x > 1 (Using Sine Wave Method: Watch this video to learn the method)
But our condition was x>=0
=> Intersection will be x > 1



Case 2: Whatever is inside the Absolute Value < 0
x < 0
=> |x| = -x (Watch this video to learn more about Basics of Absolute Value)
=> -x < \(x^2\)
=> \(x^2\) + x > 0
=> x*(x+1) > 0



=> x < -1 or x > 0 (Using Sine Wave Method: Watch this video to learn the method)
But our condition was x < 0
=> Intersection will be x < -1

So, x > 1 or x < -1
=> x^2 > 1

So, Answer will be E
Hope it helps!

Watch the following video to learn the Basics of Absolute Values


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i tried plugging in values to eliminate options.
can someone tell me how which values i can pick for option C to eliminate it

can someone tell me how which values i can pick for option C to eliminate it Bunuel KarishmaB

Hello ShaunakSawant,

Let me begin by telling you that this approach of plugging in values is not scalable (would take a lot of time to check each choice) and something that I would not recommend. Had you adopted the methodical approach of solving the question, you would have known exactly what value would eliminate choice C.

That said, I will address your request right now. 😊

You wished for a value to help us eliminate option C: x > 1. If we find a case where the original condition, |x| < x\(^2\) is true but condition x > 1 is not true, then we will be done - that is, we will be able to reject choice C as need not be true.

Now, since we want x > 1 to not be true, our x must be anywhere in the range x <= 1. And remember that it must satisfy the condition given in the question: |x| < x\(^2\).

To find such numbers, let’s just take numbers from the range x <= 1 and plug them into the given inequality: |x| < x\(^2\)to see which ones work. (Again, the ones that do work will help us reject choice C.)

1. x = 1
   |x| = 1 and x\(^2\) = 1
   |x| < x\(^2\) – NOT TRUE
2. x = 0
   |x| = 0 and x\(^2\) = 0
   |x| < x\(^2\) – NOT TRUE
3. x = -1
   |x| = 1 and x\(^2\) = 1
   |x| < x\(^2\) – NOT TRUE
4. x = -2
   |x| = 2 and x\(^2\) = 4
   |x| < x\(^2\) – TRUE

So, we have found a value in the range x <= 1 that satisfies the original condition! Hence, choice C can be rejected.

Note: I would again suggest that you build methodical approaches to solve questions. This will enable you to tackle the hardest of questions with ease.
You can check the solution given by BrushMyQuant just above your query.

Hope this helps!

Best Regards,
Ashish Arora
Quant Expert, e-GMAT
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