Is |x - 1|

Question

Is  |x - 1| 0

masuhari · Answer

expanding #1, x-1 <= 1 OR x <= 2, which doesn't help us solve is Mod(x-1) < 1. So A and D are out.

expanding #2, x^2 > 1 or x > 1, again not sufficient enough to answer Mod(x-1) < 1. So B is out.

But the 2 statements together, we get 1 < x <= 2, again not sufficient to say whether Mod(x-1) < 1. So my choice is E

amitdgr · Answer

|x - 1| < 1 has two cases case 1) x - 1 is positive, then x - 1 < 1 ===> x < 2 case 2) x - 1 is negative, then -(x - 1)<1 ===> x-1 > -1 ===> x>0 so the question is asking if x>0 and x<2, in other words does x lie in between 0 and 2 ( 0 and 2 NOT included) when we have x²<1 --- we write it as |x|<1 so stmt1) (x - 1)^2 <= 1 ===> |(x - 1)| <= 1 case 1) (x - 1) is positive, then x-1<=1 or x <=2 case 2) (x-1) is negative, then -(x-1)<=1 or x-1 >= -1 ====> x>=0 this means x>=0 and x<=2, in other words x lies in between 0 and 2 ( here 0 and 2 are INCLUDED, because we have "=" sign). This is NOT sufficient to say whether x lies between 0 and 2, with 0 and 2 not included. Stmt2) x^2 - 1 > 0 ===> x^2 > 1 ===> |x|>1 case 1) x is positive , then x >1 case 2) x is negative, then -x>1 ===> x < -1 Again, not sufficient. X can be between 0 and 2 and can be out of the range. (1) and (2) COMBINED When we have such problems, we combine the range we get from the 2 statements and check if the overlapping range lies within the range which the question is asking for. statement (1) says x lies between 0 and 2 (both inclusive) statement (2) says x lies outside -1 and 1. the overlapping range would be x>1 and x<=2. still this info is not sufficient to say whether x lies in the range with 2 not included. Since we have <=2, x could be = 2 and thus not satisfy. Other values of x >1 and <2 satisfy. Not sufficient. so E

scthakur · Answer

E.

From stmt1: |x-1| <= 1 and is not sufficient as |x-1| can also equal 1.
From stmt2, |x-1| > 0 and is not sufficient as |x-1| can also be > 1.

Combining two, also, insufficient.

amitdgr · Answer

scthakur .. we can take x²-1 > 0 as |x-1|>0 ??

amitdgr · Answer

i guess scthakur got it wrong then .....

scthakur · Answer

(x^2-1) > 0 means (x-1)(x+1) > 0. This means, either (x-1) is >0 or (x-1) 0.

I hope, I am clear.

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

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