In DS questions, the idea is to break down the question stem to a situation that can be dealt with easily. In this question, the question stem may appear a bit too short for your liking. However, you can still break it down to a simpler level.
|x|<1 is satisfied by the range -1<x<1. So, in other words, the question is asking us to find out if -1<x<1. This is what I meant about breaking down the stem. Now that you have done this, you know what to look for in the statements.
Also remember that |x| = √(\(x^2\) ).
Therefore, in statement I, we can say, |x+1| = √(〖\((x+1)〗^2\)) and |x-1| = √(\(〖(x-1)〗^2\) )
So, √(〖\((x+1)〗^2\) ) = 2 √(\(〖(x-1)〗^2\) ). On squaring both sides and simplifying, we get,
3\(x^2\) – 10 x + 3 = 0. Solving this quadratic equation, we get x = 3 or x = 1/3.
This is true. The distance of the number 3 from -1 is indeed double that of its distance from 1. Similarly, the distance of the number 1/3 from -1 is double that of its distance from 1. So, it actually ties in with the definition of |x-a| on which statement I is framed.
However, all this does not still give us a unique value of x. Statement I is insufficient.
Answer options A and D can be eliminated. Possible answer options are B, C or E.
Statement II alone says x is not equal to 3. This is clearly insufficient to determine if -1<x<1. Answer option B can be eliminated.
Combining both statements, we know that x has to be equal to 1/3 since it cannot be equal to 3. The combination of statements is sufficient. Answer option E can be eliminated.
The correct answer option is C.
Hope this helps!
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