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Is x > 1?

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Re: Is x>1? (1) (x+1)(|x|-1) > 0 (2) |x|<5 [#permalink] New post 11 Jun 2014, 18:39
sevaro wrote:
Is x>1?

(1) (x+1)(|x|-1) > 0
(2) |x|<5

I got it right plugging in number. Any other options?

Thanks

This question is rated as hard by GMAC.


ST1: When x > 0 we have: (x+1)(x-1) > 0 then x^2 - 1 > 0 --> x^2 > 1 ==> x < -1 or x > 1, because x > 0 then x > 1
When x < 0 we have: -(x+1)(x+1) > 0 ~ - (x+1)^2 > 0 not existed.
SUFFICIENT.
ST2: -5 < x < 5. INSUFFICIENT.
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Re: Is x>1? (1) (x+1)(|x|-1) > 0 (2) |x|<5 [#permalink] New post 11 Jun 2014, 23:44
Expert's post
sevaro wrote:
Is x>1?

(1) (x+1)(|x|-1) > 0
(2) |x|<5

I got it right plugging in number. Any other options?

Thanks

This question is rated as hard by GMAC.


Question: Is x > 1?

(1) (x+1)(|x|-1) > 0
For the left hand side to be positive, either both factors are positive or both are negative.

If both are positive,
x+1 > 0, x > -1
AND
|x|-1 > 0, |x|> 1 which means either x < -1 or x > 1
This is possible only when x > 1

If both are negative,
x+1 < 0, x < -1
AND
|x|-1 < 0, |x| < 1 which means -1 < x < 1
Both these conditions cannot be met and hence this is not possible.

This gives us only one solution: x > 1
So we can answer the question asked with "Yes".

(2) |x|<5
This implies that -5 < x < 5
x may be less than or more than 1. Not sufficient.

Answer (A)
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Re: Is x > 1? [#permalink] New post 12 Jun 2014, 02:10
enigma123 wrote:
Is x > 1?

(1) (x+1) (|x| - 1) > 0

(2) |x| < 5


You have to remember Z O N E D (Zero, One, Negative, Extremes, Decimals)

Statement I is sufficient:

We cannot plug in zero and 1 as the expression (x+1) (|x| - 1) will not hold true. All numbers greater than 1 will hold true for the expression. All decimals and negative numbers will make the expression negative.

Hence the value of x will always be greater than 1.

Statement II is insufficient:

x = 4 (YES) and x = -2 (NO)

Hence the answer is A
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Re: Is x>1 [#permalink] New post 13 Jun 2014, 09:55
Bunuel wrote:
Good question. +1.

Is x> 1?

(1) (x+1) (|x| - 1) > 0. Consider two cases:

If x>0 then |x|=x so (x+1) (|x| - 1) > 0 becomes (x+1) (x - 1) > 0 --> x^2-1>0 --> x^2>1 --> x<-1 or x>1. Since we consider range when x>0 then we have x>1 for this case;

If x\leq{0} then |x|=-x so (x+1) (|x| - 1) > 0 becomes (x+1) (-x - 1) > 0 --> -(x+1) (x+1) > 0 --> -(x+1)^2>0 --> (x+1)^2<0. Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that (x+1) (|x| - 1) > 0 holds true only when x>1. Sufficient.


(2) |x| < 5 --> -5<x<5. Not sufficient.

Answer: A.

Hope it's clear.


Thanks Bunnel.
I have a question regarding the modulus of X.
|X| = -X when X<= 0
|X| = X when X> 0
Which range does zero belong to? Since we know the result in the end solely depends on whether the absolute value of x (whether it is positive or negative) and zero is rather inconsequential in deciding whether |X| is to become -X or +X.
So, is
|X| = -X when X< 0
|X| = X when X >= 0 wrong? (if you observe I have swapped the equal to zero sign from the negative to the positive case)

So where exactly does zero fall in a modulus scenario? It is important because it decides whether the values on the verge of the range are considered in the solution set or not.
Hope the description of my question is clear enough!
Thank you in advance!

My guess is that we will need to plug in the values on the edges of the range and then decide where the zero falls (or rather which values are to be considered)
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Re: Is x>1 [#permalink] New post 13 Jun 2014, 10:03
Expert's post
Kconfused wrote:
Bunuel wrote:
Good question. +1.

Is x> 1?

(1) (x+1) (|x| - 1) > 0. Consider two cases:

If x>0 then |x|=x so (x+1) (|x| - 1) > 0 becomes (x+1) (x - 1) > 0 --> x^2-1>0 --> x^2>1 --> x<-1 or x>1. Since we consider range when x>0 then we have x>1 for this case;

If x\leq{0} then |x|=-x so (x+1) (|x| - 1) > 0 becomes (x+1) (-x - 1) > 0 --> -(x+1) (x+1) > 0 --> -(x+1)^2>0 --> (x+1)^2<0. Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that (x+1) (|x| - 1) > 0 holds true only when x>1. Sufficient.


(2) |x| < 5 --> -5<x<5. Not sufficient.

Answer: A.

Hope it's clear.


Thanks Bunnel.
I have a question regarding the modulus of X.
|X| = -X when X<= 0
|X| = X when X> 0
Which range does zero belong to? Since we know the result in the end solely depends on whether the absolute value of x (whether it is positive or negative) and zero is rather inconsequential in deciding whether |X| is to become -X or +X.
So, is
|X| = -X when X< 0
|X| = X when X >= 0 wrong? (if you observe I have swapped the equal to zero sign from the negative to the positive case)

So where exactly does zero fall in a modulus scenario? It is important because it decides whether the values on the verge of the range are considered in the solution set or not.
Hope the description of my question is clear enough!
Thank you in advance!

My guess is that we will need to plug in the values on the edges of the range and then decide where the zero falls (or rather which values are to be considered)


You can include 0 in either of the ranges:

When x\leq{0} then |x|=-x, or more generally when some \ expression\leq{0} then |some \ expression|={-(some \ expression)}. For example: |-5|=5=-(-5).

When x\geq{0} then |x|=x, or more generally when some \ expression\geq{0} then |some \ expression|={some \ expression}. For example: |5|=5.

The point is that |0|=0, so it doesn't matter in which range you include it.

P.S. BTW your question was already answered in this very thread: is-x-134652.html#p1261810
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