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Is |x-1| < 1?

This topic has 1 expert reply and 0 member replies

Is |x-1| < 1?

Post Sun Feb 11, 2018 3:36 am
Is |x-1| < 1 ? $$(1)\ \ (x-1)^2>1$$ $$(2)\ \ \ x<0$$ The OA is the option D.

I see that the statement (1) is sufficient. But, how can I get an answer using the statement (2)? I need your help experts.

Thanks in advanced.

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DavidG@VeritasPrep Legendary Member
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Post Thu Feb 15, 2018 6:15 pm
M7MBA wrote:
Is |x-1| < 1 ? $$(1)\ \ (x-1)^2>1$$ $$(2)\ \ \ x<0$$ The OA is the option D.

I see that the statement (1) is sufficient. But, how can I get an answer using the statement (2)? I need your help experts.

Thanks in advanced.
You can evaluate statement 2 using the number line and a little logic. Remember that the absolute value is the distance from 0 on the number line. If you were start at (-1), you'd be 1 unit from 0. If you were to add a negative number to (-1), you'd move further to the left of 0, thus increasing the absolute value to something greater than 1. That's effectively what's happening in statement 2 because x is negative. Put another way, you can think of the original question as |(-1) + x| < 1? So we're adding x, which is negative, to -1, thus producing an expression whose absolute value will always be greater than 1.

(And you can see this by picking numbers as well. If x = -2, then |-2 -1| = |-3| = 3. Clearly this is not less than 1, so we get a NO. No matter what we pick for x, the value within the absolute value brackets will be to the left of one, meaning that the absolute value - or distance from 0 -will be greater than 1.)

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