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

Hi Krishma,
the link that you posted is probably best for summary. It ends with the comment "Anyone wants to add ABSOLUTE VALUES....That will be a value add to this post ". Can you suggest any such thread where absolute value side is summarized like this as well.
Appreciate your input
Thanks

Please check my signature for the link to the relevant blog posts.

Hi bunuel,
sorry , but I didn't undestrand how you reduce |x|<x^2 by |x| to get the solution...x<-1 and x>1
please, could you explain me in more details?
Thanks

Express \(x^2\) as \(|x|*|x|\), so we have that: \(|x|<|x|*|x|\) --> reduce by \(|x|\) (notice that \(|x|\) is positive, thus we can safely divide both parts of the inequality by it) --> \(1<|x|\) --> \(x<-1\) or \(x>1\).

Hope it's clear.

Got the answer from : if-x-x-2-which-of-the-following-must-be-true-99506-20.html#p1098244

But can some one draw the graph

|X| < x^2 means X is not a decimal. means X < -1 or x > 1. Square of anything will be positive. Hence. A
Hope it helps.

|x| will be less than x^2 in the region where the graph of |x| is below the graph of x^2.
Graph of y = |x| is a V at (0, 0) and graph of y = x^2 is an upward facing parabola at (0, 0)



Just looking at the graph should tell you that answer must be (A). You don't need to solve. x will lie in one of two ranges x > a or x < -a. II and III will not work in any case. I must work since one of them has to be true according to the options.

1. |x|<x^2 =>

(i) x < x^2 or
(ii) -x < x^2

2. (1) => x>1 or x<-1 . Only this satisfies both (i) and (ii)

Is x^2>1 always true? Yes follows from (2)
Is x>0 always true, No because x could be <-1
Is x<-1 always true?, No because x could be >1

So answer is choice A.

Here's the way I look at it:

I think were all in agreement that the conditions for the statement is x<-1 and x>1. From this we can conclude this statement is true if we have values of x<-1 or x>1. But for all values to be true if I plug a value of x<-1 or x>1 into the answer choice inequalities then the inequality must hold up.

So looking at my option choices:

First option: x^2 > 1. Let's pick a number less than -1. Say -2. Is (-2)^2 > 1? Yes. Now let's look for a number x>1. Say 2. Is (2)^2>1. Yes. So this holds true for all of our conditions of x (x>1, x<-1).

Second option: x>0. Let's pick the same numbers we picked before. 2,-2. Is 2>0. Yes. Is -2>0. No this won't work if x<-1 or x is between 0 and 1.

Third option: x<-1. Again let's use the same numbers that we have been using to validate all possible solutions. Say x is 2. Is 2<-1? No. Is -2<-1? Yes. However both don't satisfy the equation

The reason why the first one does is because of the x^2. When were squaring both a positive and negative number that is x<-1 or x>1 then the result will always be a positive number that is greater than one.

Bunuel does this interpretation make sense?

Posted from my mobile device

Yes, that's correct.

From the stem we know that x is some number either less than -1 or more than 1.

Which of the statement is true about that number?

I says that the square of that number is greater than 1. Well, square of ANY number less than -1 or more than 1 is greater than 1. So, no matter what x actually is, this statement is always true.

II says that that number is greater than 0. That might not be true if it's less than -1.

III says that that number is greater less than -1. That might not be true if it's more than 1.

So, only statement I is true.

Hope it's clear.

Hi Bunuel,

I have a small doubt. Can we solve the obve equation like this:

if x>0,

\(x^2 > x\) (now dividing both the sides by x)

we get x > 1 (this is one range)

the other range is when x <0 and |x| =-x,

we get \(x^2 > -x\) (dividing by x)

x>-1

so the range is x>1 and x>-1.

where am i going wrong?

x>1 and x>-1 does not make any sense.

As for the mistake you are making, when dividing x^2 > -x by x, which is negative, you should flip the sign: x < -1.

Hi,
i wonder can we solve this inequality like this square each side
thus we would have x^2<x^4 or X^2(x-1)(x+1)<0, thus we have following solution x (-1:0) and (0:1).
So based on that NONE is the answer.
Please correct me or am i making some fundamental error.

x^2 < x^4
x^2 - x^4 < 0
x^2 (1 - x^2) < 0
x^2(x^2 - 1) > 0 (multiplying by -1 and hence flipping the inequality sign)
x^2(x - 1)(x + 1) > 0

The solution to this is x < -1 or x > 1.
You ignore x^2 because it has even power. The transition points are only -1 and 1.

Check the wavy line method here: https://youtu.be/PWsUOe77__E

x^2 < x^4

Since the power is even both sides so both sides must be positive which has two inferences
1) The range of Negative values of x will be valid along with the same range of positive values of x
2) Higher powers of x are greater than lower powers of x only when x is greater than 1 (for positive values of x)

I.e. the Required range must be x >1 or x < -1

ALTERNATIVELY

We can solve the inequation,
x^2 - x^4 < 0
x^2 (1 - x^2) < 0
x^2(x^2 - 1) > 0 (multiplying by -1 and hence flipping the inequality sign)
x^2(x - 1)(x + 1) > 0

Since x^2 is positive for all values of x except zero therefore

(x - 1)(x + 1) > 0
I.e. both the parts must be either positive or both must be Negative

For both of (x-1) and (x+1) to be positive, x must be greater than 1
For both of (x-1) and (x+1) to be negative, x must be less than -1
The solution to this is x < -1 or x > 1.

Hi,


Doubt regarding third statement.

The expression in the question is valid for all x>1 and X<-1
As per this even third option should be right.

Also, please tell me how to identify these options as in some questions, a part range is also among the answers while in others it is not. I always get confused in these questions. Is it because of "must" or "could"? Could you please mention similar questions where I can practice for this?

"Must" Means It will be true for all possible values of x
"Could" Means It may be true for some values of x and may not be true for other values of x

About the third:
III. x<-1
Take some values of x less than -1 and check if the Inequation is satisfied
@x = -2, |x|<x^2 this is satisfied
@x = -3, |x|<x^2 this is satisfied as well
for all values of x less than -1
|x|will have positive and smaller absolute value than absolute values of x^2 which will be positive.
x less than -1 is an acceptable range of values of x but since x may be positive (greater than 1) as well so it is not necessary that x be less than -1

I hope this helps!