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

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boomtangboy
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If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink] New post   Updated on: 09 Jun 2019, 22:05
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If x ≠ 0, is |x| < 1 ?

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

Originally posted by boomtangboy on 08 Apr 2012, 09:43.
Last edited by Bunuel on 09 Jun 2019, 22:05, edited 3 times in total.
Renamed the topic and edited the question.
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If x 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink] New post   08 Apr 2012, 10:01
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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.
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Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink] New post   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|)

Thanks!
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Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink] New post   19 Apr 2012, 13:25
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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
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Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink] New post   19 Apr 2012, 21:44
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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 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).
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Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink] New post   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] New post   07 Jul 2013, 18:54
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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?
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Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink] New post   07 Jul 2013, 23:18
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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.

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

Do I make myself clear btw?
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Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink] New post   11 Sep 2016, 03:21
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iliavko wrote:
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?

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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If x 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink] New post   23 Nov 2018, 19:33
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adkikani wrote:
If \(x\neq{0}\), is \(|x|\)< 1?

(1) \(x^2\) < 1

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


If \(x\neq{0}\), is \(|x|\)< 1?
This means -1<x<1, where \(x\neq{0}\)

(1) \(x^2\) < 1
\(x^2<1\) means |x|<1, since only fractions with absolute value less than 1 will have their square less than 1.
Sufficient

(2) \(|x|\) < \(1/x\)
Since |x| >0, we can write 1/|x|x>1
Since |x| and 1 are positive, x also must be positive for above to be true.
And since X is also positive, |X|=X, and X<1/X can be written as x^2<1
Therefore X is between 0 and 1
Sufficient

D­
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Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink] New post   09 Jun 2019, 12:41
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Bunuel wrote:
iliavko wrote:
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?

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.


Can you please elaborate on how a "partial" range in 2) (0 < x < 1) satisfies the full range that is required by the original prompt? I would think that the ranges have to be exact same, i.e. 1) provides a range that is exact same as compared to that in the prompt.

Thanks.
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Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink] New post   09 Jun 2019, 22:07
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mykrasovski wrote:
Bunuel wrote:
iliavko wrote:
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?

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.


Can you please elaborate on how a "partial" range in 2) (0 < x < 1) satisfies the full range that is required by the original prompt? I would think that the ranges have to be exact same, i.e. 1) provides a range that is exact same as compared to that in the prompt.

Thanks.


Any x, which is between 0 and 1, will obviously fall into the range between -1 and 1, so we can answer the question with an YES: x IS between -1 and 1.
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Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink] New post   12 Oct 2019, 16:47
This is what I did for statement 2:

|x|<1/x
As |x| will always be non negative, so the equation becomes:
X<1/x
Multiplying both sides by x
X^2<1
X^2-1<0 which is then to be worked out as the same way for st 1.

Experts please let me know is this method correct? Bunuel chetan2u VeritasKarishma
Please help

Posted from my mobile device
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If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink] New post   12 Oct 2019, 19:43
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Shef08 wrote:
This is what I did for statement 2:

|x|<1/x
As |x| will always be non negative, so the equation becomes:
X<1/x
Multiplying both sides by x
X^2<1
X^2-1<0 which is then to be worked out as the same way for st 1.

Experts please let me know is this method correct? Bunuel chetan2u VeritasKarishma
Please help

Posted from my mobile device


Yes Shef08, you can do this here because \(\frac{1}{x}>|x|\) means \(\frac{1}{x}\geq{0}\), which in turn means \(x\geq{0}\), so you can cross multiply..

exc4libur , as we can see BOTH sides are non-negative, we can square both sides to get our answer ..
\(\frac{1}{x}>|x|\)......\(\frac{1}{x^2}>|x|^2....x^4<1\)
Now x>0, so 0<x<1
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Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink] New post   13 Oct 2019, 05:44
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Shef08 wrote:
This is what I did for statement 2:

|x|<1/x
As |x| will always be non negative, so the equation becomes:
X<1/x
Multiplying both sides by x
X^2<1
X^2-1<0 which is then to be worked out as the same way for st 1.

Experts please let me know is this method correct? Bunuel chetan2u VeritasKarishma
Please help

Posted from my mobile device


If we write |x| < 1/x as x < 1/x, then we are assuming that x is positive because |x| = x only when x >= 0.

x < 1/x
I cannot multiply both sides by x without knowing the sign of x. Here x could be positive or negative e.g.
when x = 1/2, x < 1/x
when x = -2, x < 1/x

So I won't multiply both sides by x. I take 1/x to the other side.

x - 1/x < 0

(x^2 - 1)/x < 0

So 0 < x < 1 or x < -1
But x must be positive so only 0 < x < 1 satisfies.

Now try putting |x| = -x when x is negative.
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Re: If x 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink] New post   01 Jul 2022, 03:06
boomtangboy wrote:
If x ≠ 0, is |x| < 1 ?

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


A Masterpeice from OG

By working on the Question stem, we will have -1<x<1,

S1. x^2 < 1, it means -1<x<1. Sufficient
S2. |x| < 1/x, it means x has to be positive, because | | is always positive and it is less than a value. So x has to be positive.
Now as x is positive, and the reciprocal is greater than a number, it means the number has to be between 0 & 1. Sufficient.

Hence D, as both statement alone is sufficient
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Re: If x 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink] New post   12 Feb 2023, 10:39
[quote="KarishmaB"]

KarishmaB

Hi Karishma 🙏

please advise in x-1/x<0

(X^2-1)/x<0
(x-1)(x+1)/x<0
Now x cannot be zero. As denominator can not zero

But how do we get

0 < x < 1 or x < -1

I understand that x must be positive

Posted from my mobile device
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Re: If x 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink] New post   12 Feb 2023, 21:48
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puneetfitness wrote:
KarishmaB wrote:

KarishmaB

Hi Karishma 🙏

please advise in x-1/x<0

(X^2-1)/x<0
(x-1)(x+1)/x<0
Now x cannot be zero. As denominator can not zero

But how do we get

0 < x < 1 or x < -1

I understand that x must be positive

Posted from my mobile device


Check out this post:
https://gmatclub.com/forum/inequalities ... 91482.html

Let me know if you have any doubts in it.
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Re: If x 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
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