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

1

This post was BOOKMARKED

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.

Re: Is x>1? (1) (x+1)(|x|-1) > 0 (2) |x|<5 [#permalink]
11 Jun 2014, 23:44

1

This post received KUDOS

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.

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 _________________

Perfect Scores

If you think our post was valuable then please encourage us with Kudos

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)

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 _________________

Can we try this question by evaluating each of these expressions separately? Is this a correct approach, though time consuming?

1) x+1 > 0 2) x-1 > 0 3) -x-1 > 0

If correct, could you please demonstrate on how to proceed?

This would not the best approach. You should consider two cases for |x|: when x<0, then |x| = x and when x>0, then |x| = -x. As well as two cases when (x+1) and (|x| - 1) are both positive and both negative. _________________

Can we try this question by evaluating each of these expressions separately? Is this a correct approach, though time consuming?

1) x+1 > 0 ---> x > -1 2) x-1 > 0 ---> x > 1 3) -x-1 > 0 ---> -x > 1 or x > -1

If correct, could you please demonstrate on how to proceed?

This would not the best approach. You should consider two cases for |x|: when x<0, then |x| = x and when x>0, then |x| = -x. As well as two cases when (x+1) and (|x| - 1) are both positive and both negative.

Fair enough Bunuel, but just for the sake of understanding and learning the concept.

If we proceed with the 3 equations, I am getting the results in red above. Comparing them, we can take the greatest limiting factor, which is x>1, and prove sufficiency. I find this approach a little easier. Is my arithmetic correct?

Can we try this question by evaluating each of these expressions separately? Is this a correct approach, though time consuming?

1) x+1 > 0 ---> x > -1 2) x-1 > 0 ---> x > 1 3) -x-1 > 0 ---> -x > 1 or x > -1

If correct, could you please demonstrate on how to proceed?

This would not the best approach. You should consider two cases for |x|: when x<0, then |x| = x and when x>0, then |x| = -x. As well as two cases when (x+1) and (|x| - 1) are both positive and both negative.

Fair enough Bunuel, but just for the sake of understanding and learning the concept.

If we proceed with the 3 equations, I am getting the results in red above. Comparing them, we can take the greatest limiting factor, which is x>1, and prove sufficiency. I find this approach a little easier. Is my arithmetic correct?

Thank you.

Again you should consider 2*2 = 4 cases.

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

1. x < 0 --> (x+1)(-x - 1) > 0.

(x + 1) > 0 and (-x - 1) > 0 --> x > -1 and x < -1. No solution here.

2. x < 0 --> (x+1)(-x - 1) > 0.

(x + 1) < 0 and (-x - 1) < 0 --> x < -1 and x > -1. No solution here.

3. x > 0 --> (x+1)(x - 1) > 0.

(x + 1) > 0 and (x - 1) > 0 --> x > -1 and x > 1. This gives x > 1.

4. x > 0 --> (x+1)(x - 1) < 0.

(x + 1) < 0 and (x - 1) < 0 --> x < -1 and x < 1. No solution here.

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

(x + 1) > 0 and (-x - 1) > 0 --> x > -1 and x < -1. No solution here.

2. x < 0 --> (x+1)(-x - 1) > 0.

(x + 1) < 0 and (-x - 1) < 0 --> x < -1 and x > -1. No solution here.

3. x > 0 --> (x+1)(x - 1) > 0.

(x + 1) > 0 and (x - 1) > 0 --> x > -1 and x > 1. This gives x > 1.

4. x < 0 --> (x+1)(-x - 1) < 0.

(x + 1) < 0 and (-x - 1) < 0 --> x < -1 and x > -1. No solution here.

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

Hope it's clear.

Hi Bunuel,

Kindly pardon me. I am unable to understand your explanation. Could you please explain how you arrived at the 4 equations? I think the 4 cases have come up because x could be either positive or negative. Is it so?

Further you've mentioned that we've to consider 2x2 cases. I understand that you have taken 3 cases where x < 0 and 1 case where x > 0. I'm totally lost now.

(x + 1) > 0 and (-x - 1) > 0 --> x > -1 and x < -1. No solution here.

2. x < 0 --> (x+1)(-x - 1) > 0.

(x + 1) < 0 and (-x - 1) < 0 --> x < -1 and x > -1. No solution here.

3. x > 0 --> (x+1)(x - 1) > 0.

(x + 1) > 0 and (x - 1) > 0 --> x > -1 and x > 1. This gives x > 1.

4. x < 0 --> (x+1)(-x - 1) < 0.

(x + 1) < 0 and (-x - 1) < 0 --> x < -1 and x > -1. No solution here.

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

Hope it's clear.

Hi Bunuel,

Kindly pardon me. I am unable to understand your explanation. Could you please explain how you arrived at the 4 equations? I think the 4 cases have come up because x could be either positive or negative. Is it so?

Further you've mentioned that we've to consider 2x2 cases. I understand that you have taken 3 cases where x < 0 and 1 case where x > 0. I'm totally lost now.

Please help!

I had typos there. Edited now. Anyway the point is that we consider two cases for |x| and then two cases for the multiples both negative and both positive. _________________

I had typos there. Edited now. Anyway the point is that we consider two cases for |x| and then two cases for the multiples both negative and both positive.

Thanks Bunuel. Understand it now much better.

Reading your post along with this one is-x-134652-20.html#p1373275 from VeritasPrepKarishma helped. _________________

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.

Hi Bunuel

Why are we not changing the inequality sign in statement 1 when we assume x is negative-ideally we should. In that case, we get x<-1 from -(x+1)^2<0 as one of (x+1) can be eliminated.

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.

Hi Bunuel

Why are we not changing the inequality sign in statement 1 when we assume x is negative-ideally we should. In that case, we get x<-1 from -(x+1)^2<0 as one of (x+1) can be eliminated.

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

Add (x+1)^2 to both sides: \(0>(x+1)^2\), which is the same as \((x+1)^2<0\). _________________

For both of the parts to be positive we can see that x >1 . Just by trying few values you can figure this out. X cant be Zero as then the second part becomes - . X cant be 1 as then second part becomes 0 and hence the whole LHS becomes Zero.

For both of the parts to be negative we try any value less an Zero and see that no value will satisfy the equation. Hence X cannot be negative.. Hence A is Sufficient.

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