modulus

[gmatnmein2010]

Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

[shashank.ism]

Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

for |x-1|<1 --> -1<x-1<1 --> 0<x<2 --- (i)
st. 1 : (x-1)^2<=1 --> -1<= x-1<= 1 --> 0<=x<=2 x= 0, 2 out of range .............not sufficient
st. 2: x^2>1 --> -1<x<1 -<x<=0 out of range .............not sufficient
combined /: x=0 is out of range .............not sufficient
Ans E

[gmatmachoman]

Agreed E

[gmatmachoman]

Agreed E

(1) (x-1)^2 <= 1

--->

(x-1) <=1
-->x<=2

Let us try toplug in some values of X in |x-1| < 1

If x=2 --->|x-1| < 1---> |2-1|=1

if x=-2--->|x-1| < 1--->|-2-1|=3>1...

It is not sufficeinet t say distinctly whether |x-1| < 1 using st 1


(2) x^2 - 1 > 0

--> x^2=1
--> x= +1 or-1

Plug in the values of x in |x-1| < 1

Not sufficient...

SO E

[fibbonnaci]

Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

1) (x-1) <= 1 or (x-1) >= -1

x<= 2 or x>= 0
therefore 0<= x <= 2

when x= 1 then x-1 = 0 which is less than 1
whereas when x= 2 , x-1 = 1 which is not less than 1.

Not sufficient.

2) x^2 > 1
so it means x >1 or x < -1
say x= 1.5 then mod(x-1) = 0.5 which is less than 1
whereas if x= 5 then mod(x-1)= 4 which is greater than 1.

Insufficient.

combining 1) and 2)
0<= x <= 2 and x>1 or x < -1.

so 1< x <= 2
say x = 2 then (x-1) = 1 which is equal to 1
say x= 1.5 then (x-1) = 0.5 which is less than 1.
therefore, insufficient.

IMO E.