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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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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.

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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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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01 Sep 2017, 04:20
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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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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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