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Intern
Joined: 22 Aug 2014
Posts: 42
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Is |x−1|<1 ?
(1) (x−1)^2≤1
(2) x^2−1>0
First,though it sounds absurd, try to test the condition whether sufficient or not even without using (1) or (2),. Place 5, 4<1, answer is no. Again, place -2, 3<1..still no. Now, place 2,1<1..still no.Now place 1, now we get the answer yes for 0<1. again, place 0, we get 1<1, no. So, once yes, once no suggests we need more data!
Now take (1) whether we get constant yes or constant no. By rooting, we get |x−1|≤1, means 0≤x≤2. Now we have restricted freedom to choose value. We can't choose any value less than 0 or more than 2. Lets try within this boundary. Place 0 or 2, we get 1 but is 1 less than 1?,NO. Try 1, |x−1|=0 (<1), YES,now less than 1. So, insufficient because of both outcome. We need either yes or no, not both simultaneously.
Think solely on (2) now.x^2−1>0;|x|>1. so, we have to take such values either greater than 1 or less than -1. Any less than -1 value makes |x−1| more than 1, so NO. In addition, placing any value 2 or more than 2 gives us same NO. Whatif, we try placing values within 0 to 1;say,1.5. we get |x−1|=.5(<1),so YES. Once yes,once no suggests insufficient.[if x were integer, (2) would be sufficient alone]
Now combine (1) and (2). (1) restricts the freedom to choose within 0≤x≤2. (2) restricts it by |x|>1.Even after combining, nothing changes.Still we can put 1<x≤2. Try putting 2, answer NO. Try putting 1.5, YES. So, combining still fails to give a reliable or constant answer.
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45% (01:24) correct 
