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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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)

Bunuel,

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?

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.

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!

Thanks Bunuel. Understand it now much better.

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

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

Add (x+1)^2 to both sides: \(0>(x+1)^2\), which is the same as \((x+1)^2<0\).
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Ans A: Better approach for me will be to try with some common numbers quickly. -1.5, 0.5, 0, 0.5 , 1.5

Search for a question before posting a new topic. The question would have been discussed already.

As for the question,

Per statement 1, (x+1)(|x|-1) >0

Case 1: for x \(\geq\) 0 ---> |x| = x ----> (x+1)(x-1) > 0 ----> x>1 or x<-1 but as x \(\geq\) 0 ---> only possible case is x>1

Case 2: for x<0 ---> |x|=-x ---> (x+1)(-x-1)>0 ----> \(-(x+1)^2 > 0\) ---> \((x+1)^2<0\) . Now a square can never be <0 and thus x can not be negative.

Thus the only possible case from statement 1 is for x \(\geq\) 0 which gives a definite "yes" for x>1.

Hence A is the correct answer.

Hi Bunuel,

I have a question.

(x+1)(|x|-1)=0 has x=1 and x=-1 as its solution. However, x=-1 is a repeated root. Hence, on the number line, the sign will not change when it passes x=-1.

--------(-1)---------(1)+++++++

Therefore, only when x>1, (x+1)(|x|-1)>0.

Is my solution correct? I remember my highschool math teacher said something like 'the sign doesn't change when it passes a double root'. But, it's years ago and I just want to make sure that the approach is valid.

Thank you!