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neelesh
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walker
E

\(|x - 1| < 1\) --> is x (0,2)

1. \((x - 1)^2 <= 1\) --> x e [0,2]. insufficient
2. \(x^2 - 1 > 0\) --> x e (-∞,1)&(1,+∞). insufficient
1&2 x e (1,2]. insufficient.

I think I am loosing it ....

I was able to get exactly where you finished but then I argued that using 1&2 we know that x e (1,2]....meaning |x-1| will never be less than 1. And hence we can answer the question stem as a yes or no.....That is |x-1| is actually > 1

So why should it not be C ?

Combining 1 & 2 we get x e (1,2] but this is still not sufficient since the question is really asking is x e (0,2). Picking 1.5 for instance leads to a "yes", while picking 2 leads to a "no".
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So why should it not be C ?
Because x=2 and x=1.5 satisfy both conditions, but the answers are "no" and "yes"
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chineseburned, thanks. I did not see your post
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neelesh
So why should it not be C ?
Because x=2 and x=1.5 satisfy both conditions, but the answers are "no" and "yes"


Thanks walker and chineseburned..

I think I need to take a break ....what a silly a@# mistake..... :cry:.... :wall



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