Re: Is |x-1| < 1?

Can someone explain to me how you arrive at 0<x≤2? I simplified (x-1)^2≤1 down to x(x-2)≤0 and I see that x≤2 but how do you get x<0?

Also, for #1, why couldn't x be one in addition to 0 and 2?

Thanks!