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

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If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x [#permalink] New post 08 Apr 2012, 08:43
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If x≠0, is |x| < 1 ?

(1) x^2 < 1
(2) |x| < 1/x
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Last edited by Bunuel on 08 Apr 2012, 09:02, edited 1 time in total.
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Re: If x≠0, is |x| <1? [#permalink] New post 08 Apr 2012, 09:01
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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? [#permalink] New post 19 Apr 2012, 12:25
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, 20:44
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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 23 Apr 2012, 19:41
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).


Well Explained Bunuel
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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, 00:24
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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, 04:16
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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 04 Jul 2013, 11:25
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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 04 Jul 2013, 12:00
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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 byx^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.

Sufficient.
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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, 17:54
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?
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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, 22:18
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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 08 Jul 2013, 15:39
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.

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 21 Jul 2013, 19:38
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... :|
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Re: If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x [#permalink] New post 21 Jul 2013, 20:44
Expert's post
Mountain14 wrote:
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... :|


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] New post 22 May 2014, 19:33
If |x| <1 --> -1<x<1, why not |x| < 1/x --> -1/x<x<1/x? If so, how does it lead to x>0?
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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, 22:19
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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 18 Oct 2014, 12:22
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.

ANSWER: D

boomtangboy wrote:
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] 18 Oct 2014, 12:22
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