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(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.

My answer was A... Explanation-> |x|<1 means x should be between -1 and 1.

from (1) x2-1<0 => x between -1 and 1. sufficient

from (2) if x +ve then same as 1 and sufficient if x -ve then |x|=-x -x2<1 or x2 > -1 not sufficient

so and A

Please correct me if wrong

\(x\) cannot be negative. Refer to the solution above.

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).
_________________

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

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23 Apr 2012, 20:41

Bunuel wrote:

shikhar wrote:

If x≠0, is |x| <1?

(1) x2<1 (2) |x| < 1/x

Merging similar topics.

shikhar wrote:

My answer was A... Explanation-> |x|<1 means x should be between -1 and 1.

from (1) x2-1<0 => x between -1 and 1. sufficient

from (2) if x +ve then same as 1 and sufficient if x -ve then |x|=-x -x2<1 or x2 > -1 not sufficient

so and A

Please correct me if wrong

\(x\) cannot be negative. Refer to the solution above.

Also if \(x<0\) then we have \(-x<\frac{1}{x}\) and now if we cross mulitply 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).

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

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04 Jul 2013, 05:16

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boomtangboy wrote:

If x≠0, is |x| < 1 ?

(1) x^2 < 1 (2) |x| < 1/x

Given question stem asks if |x|<1------> Is -1<x<1

from St 1 we have x^2<1 ------> -1<x<1 So Sufficient

from St2 we have

|x|<1/x

Notice that |x| is a positive value and for any Integer value |x|> 1/x -----This implies X is a fraction. In order to satisfy the above equation let us take some fractional value of x and check what happens to the above equation

x= -1/2 so we have 1/2<-2 ------No x=3/4 so we have 3/4< 4/3 ------Yes x=4/3 so we have 4/3 <3/4 ------- no

We see that when fraction is between 0<x<1 then the above equation holds true and hence x is between -1 and 1 Ans should be D....

Bunuel's solution is superb. Saves time
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Re: If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x [#permalink]

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04 Jul 2013, 12:25

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If x≠0, is |x| < 1 ?

is x<1 OR is -x<1 x>-1

Is -1<x<1?

(1) x^2 < 1 x^2<1 |x|<1

This tells us exactly what the stem looks for. SUFFICIENT

(2) |x| < 1/x is x < 1/x OR is -x < 1/x is x > -1/x SO -1/x < x < 1/x -1 < x^2 < 1 SUFFICIENT

(I am going to use Bunuel's method only because I think it makes more sense and I believe mine is wrong anyways)

|x|<1/x if |x|<1/x then 1/x MUST be positive as it is greater than an absolute value. If 1/x is positive then x must also be positive and therefore |x|<1/x is actually equal to x<1/x. Because we know that x is positive we can multiply both sides by x to simplify.

(x)* x < 1/x *(x) x^2 < 1 |x|<1 x<1 or x>-1 -1<x<1

This tells us exactly what the stem is looking for SUFFICIENT (D)

(also, could someone explain to me how to solve by the method I originally chose in #2...the one where I get the positive and negative case for |X|)

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

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04 Jul 2013, 13:00

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WholeLottaLove wrote:

If x≠0, is |x| < 1 ?

(also, could someone explain to me how to solve by the method I originally chose in #2...the one where I get the positive and negative case for |X|)

Thanks!

I.x>0

\(x<\frac{1}{x}\) As x>0, we can safely cross-multiply \(\to x^2<1 \to |x|<1.\)

II.x<0

\(-x<\frac{1}{x}\) multiply both sides by\(x^2\), which is a positive quantity \(\to -x^3<x\)\([x\neq{0}]\)

or \(x(1+x^2)>0 \to\) \((1+x^2)\) and x have same sign and as\((1+x^2)\) is always positive, thus x>0. However, this goes against our assumption. Thus, x is not negative.

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

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07 Jul 2013, 18:54

If x≠0, is |x| < 1 ?

(1) x^2 < 1 x<1, x>-1 -1<x<1

Valid X is a number that is greater than -1 (i.e. -3/4, -1/2) and less than 1 (i.e. 1/2, 3/4) either way the absolute value of any of those numbers will be less than 1. SUFFICIENT

(2) |x| < 1/x Multiply both sides by x x^2<1 Same solution as above. SUFFICIENT

(D)

Is it valid to multiply the absolute value of x on one side (like in number two) to simplify as long as x is positive?

Valid X is a number that is greater than -1 (i.e. -3/4, -1/2) and less than 1 (i.e. 1/2, 3/4) either way the absolute value of any of those numbers will be less than 1. SUFFICIENT

(2) |x| < 1/x Multiply both sides by x x^2<1 Same solution as above. SUFFICIENT

(D)

Is it valid to multiply the absolute value of x on one side (like in number two) to simplify as long as x is positive?

As, since from |x| < 1/x we concluded that x is positive, then yes we can do that: x < 1/x --> x^2 < 1.

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

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08 Jul 2013, 16:39

It helps a lot. Thanks!

Bunuel wrote:

WholeLottaLove wrote:

If x≠0, is |x| < 1 ?

(1) x^2 < 1 x<1, x>-1 -1<x<1

Valid X is a number that is greater than -1 (i.e. -3/4, -1/2) and less than 1 (i.e. 1/2, 3/4) either way the absolute value of any of those numbers will be less than 1. SUFFICIENT

(2) |x| < 1/x Multiply both sides by x x^2<1 Same solution as above. SUFFICIENT

(D)

Is it valid to multiply the absolute value of x on one side (like in number two) to simplify as long as x is positive?

As, since from |x| < 1/x we concluded that x is positive, then yes we can do that: x < 1/x --> x^2 < 1.

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

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21 Jul 2013, 20:38

Bunuel wrote:

shikhar wrote:

If x≠0, is |x| <1?

(1) x2<1 (2) |x| < 1/x

Merging similar topics.

shikhar wrote:

My answer was A... Explanation-> |x|<1 means x should be between -1 and 1.

from (1) x2-1<0 => x between -1 and 1. sufficient

from (2) if x +ve then same as 1 and sufficient if x -ve then |x|=-x -x2<1 or x2 > -1 not sufficient

so and A

Please correct me if wrong

\(x\) cannot be negative. Refer to the solution above.

Also if \(x<0\) then we have \(-x<\frac{1}{x}\) and now if we cross mulitply 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).

Bunuel,

I am not clear with this part

Also if x<0 then we have -x<\frac{1}{x} and now if we cross mulitply by negative x then we should flip the sign: -x^2>1 --> x^2+1<0

I am getting the below:

-x < 1/x ---> when we cross multiply by -ve x, we get (-x*-x ) x2 >( flipping) -1 ( -x/x) ----> X2+1 >0 ??

Pls help...
_________________

"Where are my Kudos" ............ Good Question = kudos

My answer was A... Explanation-> |x|<1 means x should be between -1 and 1.

from (1) x2-1<0 => x between -1 and 1. sufficient

from (2) if x +ve then same as 1 and sufficient if x -ve then |x|=-x -x2<1 or x2 > -1 not sufficient

so and A

Please correct me if wrong

\(x\) cannot be negative. Refer to the solution above.

Also if \(x<0\) then we have \(-x<\frac{1}{x}\) and now if we cross mulitply 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).

Bunuel,

I am not clear with this part

Also if x<0 then we have -x<\frac{1}{x} and now if we cross mulitply by negative x then we should flip the sign: -x^2>1 --> x^2+1<0

I am getting the below:

-x < 1/x ---> when we cross multiply by -ve x, we get (-x*-x ) x2 >( flipping) -1 ( -x/x) ----> X2+1 >0 ??

Pls help...

When I say "multiply by negative x", I mean multiply by x, which is negative, so simply by 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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22 May 2014, 23:19

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Hi Mahmud,

|x| <1 can be written as -1<x<1 because 1 is a constant BUT

|x| < \(\frac{1}{x}\) cannot be written as -1/x<x<1/x because 1/x is a variable

Solving,

|x| <\(\frac{1}{x}\)

RHS has to be greater than 0 (As LHS can only be +ve or 0) => \(\frac{1}{x}\) > 0 => x>0 (x cannot be -ve or 0 , Because, if x is -ve then \(\frac{1}{x}\) is -ve if x = 0 then\(\frac{1}{x}\) is not defined)

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

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18 Oct 2014, 13:22

X cannot be 0.

1. X^2 < 1 ==> |X| < 1 since X^2 cannot be negative value so both positive and negative values are possible for X. Sufficient. 2. if X=-1 then 1<-1 not possible X cannot be 0 so X > 0. X^2 < 1 same as st1. Sufficient.

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

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10 Sep 2016, 08:34

Hi Bunuel!

A question here.

How is St 2 sufficient? You say it yourself that 2) implies that 0<x<1 so that is just part of the range of the question stem ( is -1<x<1 ?) So statement two doesn't cover the negative area of the range. How can it be considered to be sufficient?

How is St 2 sufficient? You say it yourself that 2) implies that 0<x<1 so that is just part of the range of the question stem ( is -1<x<1 ?) So statement two doesn't cover the negative area of the range. How can it be considered to be sufficient?

Do I make myself clear btw?

The question asks: whether x is between -1 and 1. (2) says that x is between 0 and 1, so the answer to the question is YES.
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Re: If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x [#permalink]

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30 Dec 2016, 08:05

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