Modulus question.

Tough one...is the answer E?

|x-1| < 1

so 0<|x-1|<1
hence 0<x<2

we need to find out if x is between this range exclusively

1) (x-1)^2 <=1
x^2-2x <=0
x(x-2)<=0
testing some numbers we quickly find that this range fits the bill of 0 to 2 so 1 is suff

2) x^2-1>0
(x+1)(x-1)>0
can have +ve * +ve --> x>-1, x>1 --> x>1
or -ve * -ve --> x<-1, x<1 --> x<-1

either of these 2 answers lie both inside AND outside the desired range so insuff

Ans ought to be A

From the question stem - 0 < x < 2

From statement 1 - x could be 0 or less than zero
x could be 2 or less than 2

Then how is 'A' sufficient? what if x = 0 as 0 <= 0 is true but 0 < 0 is false.

m&m wrote:|x-1| < 1

so 0<|x-1|<1
hence 0<x<2

we need to find out if x is between this range exclusively

1) (x-1)^2 <=1
x^2-2x <=0
x(x-2)<=0
testing some numbers we quickly find that this range fits the bill of 0 to 2 so 1 is suff

2) x^2-1>0
(x+1)(x-1)>0
can have +ve * +ve --> x>-1, x>1 --> x>1
or -ve * -ve --> x<-1, x<1 --> x<-1

either of these 2 answers lie both inside AND outside the desired range so insuff

Ans ought to be A

Good call - question stem is exclusive of 0 and 2 while 1) includes them...

Since in 1) x can =0 or 2 we fall outside range in question stem.

Ans should then be E - agreed!

@@uptowngirl92 - Can you share the OA please?