Is -1 < x < 3 ? where x is a real number

(1) | |x - 1| - 1 | < 1

Let the value of |x - 1| - 1 be y
So, |y|<1, Implies -1<y<1
-1 <(|x - 1| - 1) < 1
0 < |x - 1| < 2 - Adding 1 to all the terms in the inequation.
i.e -2 < x -1 < 2
i.e -1 < x < 3 - Adding 1 to all the terms in the inequation.

So, statement 1 is sufficient to answer the question.

(2) (x + 1)(x - 3) < 0

If x < -1, then (x + 1)(x - 3)> 0.
If x > 3, then (x + 1)(x - 3)> 0.
If -1 < x < 3, then (x + 1)(x - 3) < 0.
So, If (x + 1)(x - 3) < 0, then -1 < x < 3.
So, statement 2 is sufficient to answer the question.

IMO D

Is -1 < x < 3 ?

x can be 1 or o

let x=1

then | |x - 1| - 1 | < 1
| |1- 1| - 1 | < 1

|- 1 | < 1
the answ is NO

let x=0
then | |0 - 1| - 1 | < 1
0<1
the answ is YEs

so, IMHO, (1) is out

(2) is ok. I agree with Anil

AJWILL wrote:Is -1 < x < 3 ? where x is a real number

(1) | |x - 1| - 1 | < 1

(2) (x + 1)(x - 3) < 0

the simplest would be to replace |x-1| with newly introduced parameter 'a' then

st(1) |a-1|<1 and 0<a<2. Substitute 'a' for |x-1| and solve 0<|x-1|<1 [spoiler]{if you noticed the supplied 'a' for |x-1| is similar to just solved with 'a'}[/spoiler]. Two conditions, 0<x<2 and x=!1
Since, x cannot be 1 we define the new intervals -1<x<1 and 1<x<3. Therefore st(1) is Not Sufficient.

st(2) two critical points: -1, 3; the function will increase, decrease, increase and we need to look into decrease interval only - i.e. -1<x<3 works as suitable for us. Hence, st(2) Sufficient
___+___________-____________+___
__________-1____________3__________

answer b