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absolute values and squares

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by mariofelixpasku » Fri Oct 19, 2012 10:43 am
If x≠0, is |x| < 1?
(1) x^2 < 1
(2) |x| < 1/x

Answer is: D

can we derive some formula for the first case cause it would equal x^2 - 1 < 0
hence (x-1)(x+1)<0

how do we come up with a solution?

thanks in advance
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by Jim@StratusPrep » Sat Oct 20, 2012 6:25 am
You did the problem correctly. If you test numbers from there you will see that the only place you have a negative number is between 1 and -1.
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by ines17 » Mon Feb 11, 2013 1:43 pm
I'm not sure if I understood this problem correctly. Can you please explain the solution in detail?

Thanks
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by vishalbpr » Tue Feb 12, 2013 1:08 am
In case of statement 1 you have derived to right conclusion

1) (x-1)(x+1)<0

It means -1 < x < 1, so x varies from -1 to 1, so in questions it talks about |x| which is always less than 1.

(2) |x| < 1/x
|x| is always positive, so for negative values of x this property would not hold right.
So for positive values equation is like this..

x*x - 1 < 0 when x > 0

Again x varies from 0 to 1, so 0 < x < 1, so |x| will always be less than 1.

With both the choices you can answer the question mentioned above, so D is the right answer.
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