papumba2011 wrote:pops wrote:neoreaves wrote:Is |x - 1| less than 1 ?
1). (x - 1)^2 less than and equal to 1
2). (x^2) - 1 greater than 0
1. alone:
since, (x-1)^2 <= 1
=> |x-1| <= 1
(always remember whenever you take root on both side of inequality, its absolute value that comes out i.e.
(a-b)^2 <= c^2
=> |a-b| <= |c|)
hence, yes |x-1| is less than 1!
2. alone:
since, x^2 -1 >=0
=> x^2>=1
=> either x >= 1 or x <= -1
hence, not sure whether |x-1| would be less than 1
I think we missed the '=' sign in the first statement.
If (x-1)^2 <=1
this implies |x -1|<= 1 i.e. -1<=|x-1|<=1
Since |x-1| can be equal to 1 we cannot say surely that |x-1| is less than 1. Insufficient.
From stmt 2
(x^2) - 1 > 0
Hence x must always be more than 1 or less than -1 cos of the square. As such |x-1| is always more than 1. Sufficient.
Hence answer is B.
I dont think it is B ...
(x^2) - 1 = (x+1)(x-1) > 0 --> take negative numbers x = -5 --> (-5 + 1) (-5 - 1) = (-4)(-6) = 24 > 0
we can also try x = 5 --< (5+1)(5-1) = (6)(4) = 24 Thus insufficient
Now that we know A and B are insufficient lets try C
(x - 1)^2 < = 1 ... thus (x-1) is either 1 or a fraction
and (x^2) - 1 > 0 ..Thus |x| > 1
Lets try x < 0 |x| > 1 --> lets say x = -1.1 then x - 1 = -1.1 -1 = -2.2 thus x -1 < 1 but this doesnt satisfy 1) where x- 1 has to be fraction or 1. Thus not possible
Lets try x > 0 |x| > 1 --> x = 1.1 --> x - 1 = 0.1 thus x - 1 is a fraction
Thus IMO C
