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Is |x| < 1 ?

A) \(x^4-1> 0\)

B) \(\frac{1}{1-|x|>}0\)

How to solve this algebraically ?

Hi,

To solve algebrically you first must understand the question
Is |x| < 1 ?
rewriting question statement as
Is -1<x<1?


Using first statement,

\(x^4-1> 0\)

(x^2+1)(x^2-1)>0

(x^2+1)(x+1)(x-1)>0

x^2+1 will always be greater than 1.(As x can be anything but square of anything +1 will always be greater than 1.)

So there are two points +1 and -1.

(x+1)(x-1) >0

will be solved only if x <-1 and x>1

so No.
hence Sufficient.

Using statement 2:

\(\frac{1}{(1-|x|)}>0\)

1-|x| > 0 (if denominator is +ve than only this expression can be greater than zero.)
|x|<1
so -1<x<1. Yes Sufficient.

So the answer will be D.
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Is |x| < 1 ?

(1) x^4 – 1 > 0

(2) \(\frac{1}{(1- |x|)} > 0\)


Could anyone please help with the solution.
Thanks in Advance!
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Is |x| < 1 ?

(1) x^4 – 1 > 0

(2) \(\frac{1}{(1- |x|)} > 0\)


Could anyone please help with the solution.
Thanks in Advance!

Merging topics. Please refer to the discussion above.

Also, please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Thank you.

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