Is x 1? (1) (x+1)(|x| - 1) 0 (2) |x|

Question

Is x > 1?

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

(2)  |x| < 5

Bunuel · Answer

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.

enigma123 · Answer

Thanks Bunuel. Can you please explain how did you get this?

If  x>0  then  |x|=x

Bunuel · Answer

Check this: math-absolute-value-modulus-86462.html Absolute value properties: 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 . Hope it helps.

kashishh · Answer

Dear Bunuel,
 i got (x+1)^2<0 . 
 and further solved to x+1<0 giving x <-1
 and ended up in wrong answer E. 
 square of a number cannot be negative. but just confused for (x+1)^2 where x is a variable?

Bunuel · Answer

It doesn't matter that  x  is a variable, it's still some number and so is  x+1 . So,  (x+1)^2  is a square of that number and it cannot be negative.

Also your way of solving is not correct:  (x+1)^2<0  does not mean  x+1<0  it means that  |x+1|<0 . From that you could deduce the same: since absolute value cannot be negative then this equation has no solution.

Hope it's clear.

pavanpuneet · Answer

Hi Bunuel, I understood your solution but can you please tell me where am I going wrong?

When I see the expression, (x+1)(|x|-1)>0, I immediately think that these two brackets must be either positive or negative.

Hence, if that take both of them are positive, then x>-1 and and x>1 & x<-1

For these two ranges we realize when we start plugging in number that only for x>1 the equation hold true.--------(1)

Similarly, if we take both of them to be negative, then x<-1 and x<1 & x>-1

for x<-1 for example, -2, one of the brackets turn out to be positive, which does not satisfy our initial assumption that both must be negative.

for x<1 ex 0, one bracket is +ve and one is -ve <Not Desired>; for 0, the inequality is not > 0 <not desired>

if we go still less it follows the first case

for x>-1 pretty much follows like the above. Hence nothing desired.--------(2)

Hence, x>1

A is sufficient.

Bunuel · Answer

First of all I wouldn't recommend to solve this question the way you are doing. Next, when you consider both multiples to be negative and get x<-1 from the first one, then the second multiple automatically transformes to (-x-1) , since if x<-1<0 then |x|=-x . So, we have that -x-1<0 must also be true or x>-1 , which contradicts the case for the first multiple ( x<-1 ). So, both x+1 and |x|-1 can not be negative. Also I think you got x<1 & x>-1 from |x|<1, and if yes, then it's not correct: |x|<1 means that -1

Vips0000 · Answer

Statement 1: For (x+1)(lxl-1) > 0, we should have either (x+1)>0 and (lxl-1) > 0 or (x+1)<0 and (lxl-1) < 0 when (x+1)>0 and (lxl-1) > 0 (x+1)>0 => x>-1 (lxl-1) > 0 => x>1 or x <-1 From above two, possible solution is x>1 when (x+1)>0 and (lxl-1) < 0 (x+1)<0 => x<-1 (lxl-1) < 0 => -11. which is what we wanted to ascertain. Sufficient. Statement 2: |x| <5 => -51 or not. Ans A it is.

smartyman · Answer

Please help,

When I try this question for statement 1:
(x+1) (|x| - 1) > 0
(x+1)|x|-(x+1)>0
(x+1)|x| >(x+1)
|x| > 1

x>1 ; x<-1

Bunuel · Answer

Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know its sign. So you cannot reduce both parts of inequality (x+1)|x|>(x+1) by x+1 as you don't know the sign of x+1: if x+1>0 you should write |x|>1 BUT if x+1<0 you should write |x|<1 (flip the sign). Hope it helps.

ygdrasil24 · Answer

I mean to say, we define |X| as +x when x is greater than equal to zero, and -x when x is less than zero. But I have noticed you at times have taken x as -x based on x less than equal to zero ? I mean equality cant be on both sides right ? Infact I saw you explained definition of mod function keeping equality on both +x and - x, I hope you are able to make out what doubt I have.

Bunuel · Answer

The point is that |0|=-0=0. So, in that definition we can include = sign in both cases.

Hope it helps.

tennisAndM · Answer

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.

KarishmaB · Answer

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)

PerfectScores · Answer

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

Kconfused · Answer

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 · Answer

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

AdmitJA · Answer

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?

Bunuel · Answer

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.

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

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