Is x > 1?


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

(2) \(|x| < 5\)
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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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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?

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.

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.

We need to get rid of the modulus in the expression to solve it and this is the way to do that. Check here: is-x-134652.html#p1097668

Hope it helps.
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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

Hi guys,

I just want to present an easier way to prove that I is sufficient.

For I to be positive what are the conditions?

1. x+1 > 0 this means x >-1
2.|x| -1 > 0 this means x > 1 or x < -1

then you must draw the number line ----------------[b]-1[b]-----0------1--------------> then draw this inequalities on this line and look for any unity. then you will fine x > 1

for the other side, I mean, x+1 <0 and |x| - 1 <0 you wont find any unity. so I is sufficient. Hope it clears. cheers

x can never be -2, as because you have assumed that \(x\geq0\). Also, from the inequality you have correctly arrived at, i.e. \(x^2>1\) \(\to\) x>1 OR x<-1. As assumption was \(x\geq0\). thus only x>1 condition is valid. Also,as x>1 automatically makes \(x\geq0\), thus the correct range is x>1. Sufficient.

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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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