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.
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Hello, everyone. I came across this question today in my studies and thought I would post my thoughts on Statement (1) in an effort to assist the community. (It should be quite clear that Statement (2) is NOT sufficient on its own if you have even scratched the surface of absolute value.)

To answer the question, we only need to keep in mind three questions:

1) Can x = 1?
2) Does the inequality in the statement work if x > 1?
3) Does the inequality in the statement work if x < 1?

That is it. We can use simple arithmetic to check for answers.

1) Let x = 1.

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

\((2)(0) > 0\) X

The answer to question 1 above is NO, x cannot equal 1 itself.

2) Let x = 1.1 (or something close to 1 from the greater end to check one extreme).

\(((1.1)+1)(|(1.1)| - 1) > 0\)

\((2.1)(0.1) > 0\) √

Even though the product would be close to 0, a positive times a positive will be positive, still greater than 0, no further calculation necessary. Moreover, we do not need to test larger positive numbers, since we can see that we would only end up with a larger positive product. The answer to question 2 above, then, is YES, the inequality works if x > 1.

3) Let x = 0.99 (or something close to 1 from the lesser end to check another extreme).

\(((0.99)+1)(|(0.99)| - 1) > 0\)

\((1.99)(-0.01) > 0\) X

We know that the left-hand side will be negative, so the inequality would fail. Now, there is no point in testing the unique integer 0 because we can appreciate that the positive * negative dynamic would persist. The only thing left to test is a negative value, but by this point, we do not actually need to work through anything. We can see that -1 would give us a product of 0, and that other negatives would yield the following:

a) x is between -1 and 0.
(positive) * (negative) = negative

b) x is less than -1.
(negative) * (positive) = negative

No matter what kind of number we would want to test, if x < 1, the product will always be negative, so the answer to question 3 above is NO. To recap:

1) Can x = 1? NO
2) Does the inequality in the statement work if x > 1? YES
3) Does the inequality in the statement work if x < 1? NO

We know beyond any doubt that x > 1, and the answer must be (A). Although explaining the chain of logic step by step took time, the mental process, as well as writing down numbers to ensure accuracy, took me little over a minute, and I answered confidently. If I can do it, so can you.

Good luck with your studies.

- Andrew
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Hi,
the solution here is correct, But I am not sure if it has to do with Double root in this specific instance..

Double root is generally when a polynomial has two equal roots..
so when ever you are placing the polynomial as>0, you cannot do anything to the squared part as it is always positive so it will depend on the other roots..

example--
\((x-3)^2*(x-2)>0\)..
as can be seen \((x-3)^2\)will always be >0 except at x=3.. so we require to just look for x-2..
and x-2 will be positive only whenx>2..
so solution will be x>2 given that x is NOT equal to 3..

what is happening here --
(x+1)(|x|-1)>0.
x with a value <1 will make both (x+1) and (|x|-1) of opposite sign

a) \(x<-1\)
(x+1) is negtaive and (|x|-1) is positive

b) \(-1< x<1\)
(x+1) is positive and (|x|-1) is negative

so here (x+1)(|x|-1) will be negative and there will be no solutions for (x+1)(|x|-1)>0.
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Thanks Bunuel. Can you please explain how did you get this?

If \(x>0\) then \(|x|=x\)
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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.
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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?

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.
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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.

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 \(-1<x<1\). So, again \(x<-1\) (for the first multiple to be negative) and \(-1<x<1\) (for the second multiple to be negative) cannot simultaneously be true.


Hope it's clear.
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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 => -1<x<1
Both of these can not be satisfied by any value of x.
Hence we get only 1 solution, x>1. which is what we wanted to ascertain. Sufficient.

Statement 2:
|x| <5
=> -5<x<5
Clearly not sufficient to tell whether x>1 or not.

Ans A it is.

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

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.
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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.

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

Hope it helps.
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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.

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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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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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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