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If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x

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Re: If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x  [#permalink]

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New post 05 Feb 2017, 05:34
AmritaSarkar89 wrote:
Bunuel wrote:
boomtangboy wrote:
If x≠0, is |x| <1?
(1) x2<1
(2) |x| < 1/x


If \(x\neq{0}\), is \(|x| <1\)?

Is \(|x| <1\)? --> is -\(1<x<1\) (\(x\neq{0}\))?

(1) \(x^2<1\) --> \(-1<x<1\). Sufficient.

(2) \(|x| < \frac{1}{x}\) --> since LHS (|x|) is an absolute value which is always non-negative then RHS (1/x), must be positive (as \(|x| < \frac{1}{x}\)), so \(\frac{1}{x}>0\) --> \(x>0\).

Now, if \(x>0\) then \(|x|=x\) and we have: \(x<\frac{1}{x}\) --> since \(x>0\) then we can safely multiply both parts by it: \(x^2<1\) --> \(-1<x<1\), but as \(x>0\), the final range is \(0<x<1\). Sufficient.

Answer D.


Can we solve this by plugging in numbers? I did it that way and got the correct answer. Wondering if that was just by chance or was I correct in my approach



Yes you can but you have to then validate with every kind of number i.e positive , negative , decimals , range between 0 to 1 etc.
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Re: If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x  [#permalink]

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New post 27 Aug 2017, 08:25
Bunuel mikemcgarry IanStewart shashankism Engr2012

Quote:
Also if \(x<0\) then we have \(-x<\frac{1}{x}\) and now if we cross multiply by negative \(x\) then we should flip the sign: \(-x^2>1\) --> \(x^2+1<0\) which cannot be true for any real value of \(x\) (the sum of two positive value cannot be less than zero).


why am I getting \(x^2\) > -1 for this step? I multiplied both sides of \(-x<1/x\) with -x
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If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x  [#permalink]

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New post 01 Sep 2017, 04:20
1
adkikani wrote:
Bunuel mikemcgarry IanStewart shashankism Engr2012

Quote:
Also if \(x<0\) then we have \(-x<\frac{1}{x}\) and now if we cross multiply by negative \(x\) then we should flip the sign: \(-x^2>1\) --> \(x^2+1<0\) which cannot be true for any real value of \(x\) (the sum of two positive value cannot be less than zero).


why am I getting \(x^2\) > -1 for this step? I multiplied both sides of \(-x<1/x\) with -x


You are making an algebraic mistake. If you do assume that x<0, |x| = -x; multiplying both sides of the inequality -x<1/x by -x you should get, -x*(-x) < (1/x)*(-x) ---> x^2 < -1 and this is not possible for any real 'x'. Hence, x<0 can not be true.

Additionally, you dont need to flip the signs when you multiply by -x as we are assuming that x<0, making -x >0 and when you multiply an inequality by a positive quantity, you do not need to flip the signs.

Hope this helps.
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Re: If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x  [#permalink]

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New post 15 Sep 2018, 20:35
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Re: If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x &nbs [#permalink] 15 Sep 2018, 20:35

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