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Tough inequality 1

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Tough inequality 1

by guerrero » Mon Mar 11, 2013 10:48 am
If |x|<x^2 , which of the following must be true?

I. x^2>1
II. x>0
III. x<-1

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only



This has been discussed intensively on another forum .I would like to know the approach from the experts here . I am having trouble ruling out OPTION "E" .Appreciate an elaborated solution . That will help me grasp the logic as well .

As always , thanks in advance !

OA A
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by Anju@Gurome » Mon Mar 11, 2013 10:57 am
guerrero wrote:If |x|<x^2 , which of the following must be true?
Note that, x cannot be equal to zero.

Now if x > 0, |x| = x ---> x < x² ---> x > 1
And if x < 0, |x| = -x ---> -x < x² ---> x < -1
Hence, either x < -1 or x > 1

Now,
  • I. x² > 1 ----> Always true
    II. x > 0 ----> Not always true as x can be less than -1
    III. x < -1 ----> Not always true as x can be greater than 1
The correct answer is A.
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by Brent@GMATPrepNow » Mon Mar 11, 2013 11:06 am
guerrero wrote:If |x|<x^2 , which of the following must be true?

I. x^2>1
II. x>0
III. x<-1

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
The question asks us to determine which of the statements MUST be true.
So, if we can find a case in which a statement need not be true, we can eliminate it.

Given: |x| < x^2
If this is true, what's a possible value for x?
I'd begin by trying some nice/easy values.
Does x = 0 work? No (|0| is not less than 0^2)
Does x = 1 work? No (|1| is not less than 1^2)
Does x = 2 work? Yes! (|2| is less than 2^2)
So, x could equal 2.

Check the 3 statements:
I. 2^2 > 1 True - keep statement I
II. 2 > 0 True - keep statement II
III. 2 < -1 NOT true - eliminate statement III

We can now eliminate C and E, which means the correct answer is A, B or D

Let's try some negative values for x.
Does x = -2 work? Yes! |-2| is less than (-2)^2
So, x could equal -2.

Check the 2 remaining statements:
I. (-2)^2 > 1 True - keep statement I
II. -2 > 0 NOT true - eliminate statement II

We can now eliminate B and D, which leaves us with A

Cheers,
Brent
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by GMATGuruNY » Mon Mar 11, 2013 11:35 am
guerrero wrote:If |x|<x^2 , which of the following must be true?

I. x^2>1
II. x>0
III. x<-1

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
Try to prove that I, II and III DO NOT have to be true.
Plug in easy values that satisfy the condition that |x| < x².

x=2 works, because |2| < 2².
Thus, it does not have to be true that x<-1.
Eliminate C and E, which include III.

x=-2 works, because |-2| < (-2)².
Thus, it does not have to true that x>0.
Eliminate B and D, which include II.

The correct answer is A.

Proof that I must be true:

In all cases, |x| ≥ 0.
Here, since |x| < x², |x| ≭ 0.
Thus, |x| > 0, implying that each side of |x| < x² can safely be divided by |x|:
|x| < x²
|x| < |x| * |x|
1 < |x|
1² < |x|²
x² > 1.
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by subhakam » Tue Mar 12, 2013 4:39 pm
Brent@GMATPrepNow wrote:
guerrero wrote:If |x|<x^2 , which of the following must be true?

I. x^2>1
II. x>0
III. x<-1

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
The question asks us to determine which of the statements MUST be true.
So, if we can find a case in which a statement need not be true, we can eliminate it.

Given: |x| < x^2
If this is true, what's a possible value for x?
I'd begin by trying some nice/easy values.
Does x = 0 work? No (|0| is not less than 0^2)
Does x = 1 work? No (|1| is not less than 1^2)
Does x = 2 work? Yes! (|2| is less than 2^2)
So, x could equal 2.

Check the 3 statements:
I. 2^2 > 1 True - keep statement I
II. 2 > 0 True - keep statement II
III. 2 < -1 NOT true - eliminate statement III

We can now eliminate C and E, which means the correct answer is A, B or D

Let's try some negative values for x.
Does x = -2 work? Yes! |-2| is less than (-2)^2
So, x could equal -2.

Check the 2 remaining statements:
I. (-2)^2 > 1 True - keep statement I
II. -2 > 0 NOT true - eliminate statement II

We can now eliminate B and D, which leaves us with A

Cheers,
Brent
Brent - per your response
Let's try some negative values for x.
Does x = -2 work? Yes! |-2| is less than (-2)^2
So, x could equal -2.

How and why can x NOT be a negative fraction ? if we chose x to be a negative fraction say -1/2 then answer choice A blows away. Per the question |x|< x^2 so -1/2 would still be ok and therefore x^2 will not be > than 1
Can you please explain ?
Thanks
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by subhakam » Tue Mar 12, 2013 4:43 pm
Anju@Gurome wrote:
guerrero wrote:If |x|<x^2 , which of the following must be true?
Note that, x cannot be equal to zero.

Now if x > 0, |x| = x ---> x < x² ---> x > 1
And if x < 0, |x| = -x ---> -x < x² ---> x < -1
Hence, either x < -1 or x > 1

Now,
  • I. x² > 1 ----> Always true
    II. x > 0 ----> Not always true as x can be less than -1
    III. x < -1 ----> Not always true as x can be greater than 1
The correct answer is A.
Anju - can you help me algebraically how you are getting the below please?
And if x < 0, |x| = -x ---> -x < x² ---> x < -1 ?
Per my calculation -x<x² => 0< x²+x or x²+x>0
=> x(x+1)>0 or x>0 OR x+1>0 => x>-1 - how do you get x<-1?
All help greatly appreciated!
Many thanks
Subhakam
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by Anju@Gurome » Tue Mar 12, 2013 5:17 pm
subhakam wrote:Anju - can you help me algebraically how you are getting the below please?
And if x < 0, |x| = -x ---> -x < x² ---> x < -1 ?
We know, -x < x²
Now I divided both sides by x and as x < 0, the sign of the inequality changes.
So, -x/x > x²/x ---> -1 > x ---> x < -1
subhakam wrote:Per my calculation -x<x² => 0< x²+x or x²+x>0
=> x(x+1)>0 or x>0 OR x+1>0 => x>-1 - how do you get x<-1?
No. That is wrong conclusion.
x(x + 1) > 0 means either both x and (x + 1) are positive OR both x and (x + 1) are negative. But we started with the assumption that x < 0, hence the first option is not possible here. From second option(x + 1) is negative ---> (x + 1) < 0 ---> x < -1

Hope that helps.
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by Anju@Gurome » Tue Mar 12, 2013 5:33 pm
subhakam wrote:How and why can x NOT be a negative fraction ? if we chose x to be a negative fraction say -1/2 then answer choice A blows away. Per the question |x|< x^2 so -1/2 would still be ok and therefore x^2 will not be > than 1
No, it is not ok.
x = -1/2 ---> |x| = 1/2 and x² = 1/4 ---> |x| > x²
But the question said |x| < x²

Similarly any negative fraction will not satisfy the given criteria so they are not possible values of x.

Hope that helps.
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by subhakam » Tue Mar 12, 2013 5:57 pm
Anju@Gurome wrote:
subhakam wrote:Anju - can you help me algebraically how you are getting the below please?
And if x < 0, |x| = -x ---> -x < x² ---> x < -1 ?
We know, -x < x²
Now I divided both sides by x and as x < 0, the sign of the inequality changes.
So, -x/x > x²/x ---> -1 > x ---> x < -1
subhakam wrote:Per my calculation -x<x² => 0< x²+x or x²+x>0
=> x(x+1)>0 or x>0 OR x+1>0 => x>-1 - how do you get x<-1?
No. That is wrong conclusion.
x(x + 1) > 0 means either both x and (x + 1) are positive OR both x and (x + 1) are negative. But we started with the assumption that x < 0, hence the first option is not possible here. From second option(x + 1) is negative ---> (x + 1) < 0 ---> x < -1

Hope that helps.
Dang!! Thank you for the prompt response - that helps - i complete forgot that yes x and (x+1) are in fact 2 different numbers AND 2 numbers multiplied together >0 means either both are positive or both are negative!! This case indeed both are negative! hence x<-1

Now if this was a question asking for how many solutions of x possible for |x| > x²
Then would the final answer be 4 solutions? x>1, x>0, x<0 and x<-1 ?
Please help me out here - i am trying to ensure i understand the fundamentals

Second question - is the absolute value ALWAYS positive? meaning x = -1/2 ---> |x|= +1/2 ?
Many thanks once again!
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by Anju@Gurome » Tue Mar 12, 2013 9:13 pm
subhakam wrote:Now if this was a question asking for how many solutions of x possible for |x| > x²
Then would the final answer be 4 solutions? x>1, x>0, x<0 and x<-1 ?
Please help me out here - i am trying to ensure i understand the fundamentals
I don't know what is your logic behind that conclusion so I can't point out your mistake but I'll try to explain in the same manner I did for the original problem. Before that, let me just clarify that "how many solutions" phrase is only valid when we are looking for finite number of solutions. For an inequality generally a range of values is the solution unless number of integral solution or some special condition is mentioned. And number of ranges are not same as number of solutions. Also the four ranges you mentioned are basically a single range, i.e. the whole number line split into four ranges.

To solve your problem,

If x > 0 ---> |x| = x ---> x > x² ---> 1 > x ---> 0 < x < 1
If x < 0 ---> |x| = -x ---> -x > x² ---> -1 < x ---> -1 < x < 0

Hence, the final solution set is -1 < x < 1
subhakam wrote:Second question - is the absolute value ALWAYS positive? meaning x = -1/2 ---> |x|= +1/2 ?
Yes. By definition, absolute value of a real number x is the non-negative value of x without regard to its sign. Algebraic definition of absolute value of a real number x = |x| is
  • x if x ≥ 0 and -x if x < 0
Hope that helps.
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by Anju@Gurome » Tue Mar 12, 2013 9:18 pm
guerrero wrote:If |x|<x^2 , which of the following must be true?

I. x^2>1
II. x>0
III. x<-1
Another approach to solve this problem is to exploit the fact that we can write x² = |x|*|x|
And from the given condition we know that x ≠ 0. Therefore, |x| > 0 and we can divide the both sides of the inequality with |x|.

So, |x| < x²
--> |x| < |x|*|x|
--> 1 < |x|
--> Either x < -1 or x > 1

The rest is as my first post.
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