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605-655 Level|   Absolute Values|   Inequalities|               
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puneetfitness
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If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x 13 Feb 2023, 01:17
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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[/quote][/quote]

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

Let me know if you have any doubts in it.[/quote]

KarishmaB

Hi Karishma

So this is what I got from the link you have shared with me.

From (x-1)(x+1)/x < 0

We get three points on number line ---(-1)-----0------1----

So we four ranges x>1 0<x<1 -1<x<0 and x<-1
Now if x is greater than 1 equation will be positive hence greater then
zero---> not possible

If x is between zero and 1
(x-1) is negative and x+1
positive hence equation is negative and less then 1

Say x =.5 then it becomes
(-0.5)(1.5)/0.5= -1.5


If x is between -1 to 0
Then equation becomes postive

Say x is -0.5

Then -(1.5)(0.5)/-0.5= positive hence greater then greater then 0 again not possible also x cannot be less then zero.


If x is less then -1 then equation is still negative it will (-ve)(-ve)/-ve still negative I've and less then 0

But x cannot be negative hence x between 0-1
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If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x 13 Feb 2023, 01:32
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puneetfitness
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.

KarishmaB

Hi Karishma

So this is what I got from the link you have shared with me.

From (x-1)(x+1)/x < 0

We get three points on number line ---(-1)-----0------1----

So we four ranges x>1 0<x<1 -1<x<0 and x<-1
Now if x is greater than 1 equation will be positive hence greater then
zero---> not possible

If x is between zero and 1
(x-1) is negative and x+1
positive hence equation is negative and less then 1

Say x =.5 then it becomes
(-0.5)(1.5)/0.5= -1.5


If x is between -1 to 0
Then equation becomes postive

Say x is -0.5

Then -(1.5)(0.5)/-0.5= positive hence greater then greater then 0 again not possible also x cannot be less then zero.


If x is less then -1 then equation is still negative it will (-ve)(-ve)/-ve still negative I've and less then 0

But x cannot be negative hence x between 0-1
Show more


Yes, here is an excerpt from my Inequalities book:

How to Handle an Inequality with Multiple Factors

Consider any inequality of the form (x - a)(x - b)(x - c) < 0. For clarity, let’s work with the example x(x - 3) < 0 discussed above.

Step 1: Make a number line and plot the points a, b and c on it. In our example, a = 0 and b = 3. The number line is divided into sections by these points. In our example, it is divided into 3 sections – ‘greater than 3’, ‘between 0 and 3’ and ‘less than 0.’

Step 2: Starting from the rightmost section, mark the sections with alternate positive and negative signs. The rightmost section will be positive, the next one will be negative, the next one will be positive and so on... The inequality will be positive in the sections where we have the positive signs and it will be negative in the sections where we have the negative signs.

Attachment:
Screenshot 2023-02-13 at 1.59.50 PM.png
Screenshot 2023-02-13 at 1.59.50 PM.png [ 16.58 KiB | Viewed 1355 times ]

Therefore, x*(x - 3) will be positive in the ranges ‘x > 3’ and in ‘x < 0’.
It will be negative in the range ‘0 < x < 3’.
Hence, the values of x for which x*(x - 3) < 0 is satisfied is 0 < x < 3.


Explanation:

When we plot the points on the line, the number line is divided into various sections. Values of x in the right most section will always give us positive value of the expression. The reason for this is that if x > 3, all factors will be positive i.e. x and (x - 3), both will be positive.

When we jump to the next region i.e. between x = 0 and x = 3, the values of x will give us negative values for the entire expression because now, only one factor, (x - 3), will be negative. All other factors will be positive.

When we jump to the next region on the left where x < 0, expression will be positive again because now both factors x and (x - 3) are negative. The product of two negatives is positive so the expression will be positive.
Similarly, we can solve a question with any number of factors.
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If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x 13 Feb 2023, 01:41
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KarishmaB
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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.

KarishmaB

Hi Karishma

So this is what I got from the link you have shared with me.

From (x-1)(x+1)/x < 0

We get three points on number line ---(-1)-----0------1----

So we four ranges x>1 0<x<1 -1<x<0 and x<-1
Now if x is greater than 1 equation will be positive hence greater then
zero---> not possible

If x is between zero and 1
(x-1) is negative and x+1
positive hence equation is negative and less then 1

Say x =.5 then it becomes
(-0.5)(1.5)/0.5= -1.5


If x is between -1 to 0
Then equation becomes postive

Say x is -0.5

Then -(1.5)(0.5)/-0.5= positive hence greater then greater then 0 again not possible also x cannot be less then zero.


If x is less then -1 then equation is still negative it will (-ve)(-ve)/-ve still negative I've and less then 0

But x cannot be negative hence x between 0-1


Yes, here is an excerpt from my Inequalities book:

How to Handle an Inequality with Multiple Factors

Consider any inequality of the form (x - a)(x - b)(x - c) < 0. For clarity, let’s work with the example x(x - 3) < 0 discussed above.

Step 1: Make a number line and plot the points a, b and c on it. In our example, a = 0 and b = 3. The number line is divided into sections by these points. In our example, it is divided into 3 sections – ‘greater than 3’, ‘between 0 and 3’ and ‘less than 0.’

Step 2: Starting from the rightmost section, mark the sections with alternate positive and negative signs. The rightmost section will be positive, the next one will be negative, the next one will be positive and so on... The inequality will be positive in the sections where we have the positive signs and it will be negative in the sections where we have the negative signs.

Attachment:
Screenshot 2023-02-13 at 1.59.50 PM.png

Therefore, x*(x - 3) will be positive in the ranges ‘x > 3’ and in ‘x < 0’.
It will be negative in the range ‘0 < x < 3’.
Hence, the values of x for which x*(x - 3) < 0 is satisfied is 0 < x < 3.


Explanation:

When we plot the points on the line, the number line is divided into various sections. Values of x in the right most section will always give us positive value of the expression. The reason for this is that if x > 3, all factors will be positive i.e. x and (x - 3), both will be positive.

When we jump to the next region i.e. between x = 0 and x = 3, the values of x will give us negative values for the entire expression because now, only one factor, (x - 3), will be negative. All other factors will be positive.

When we jump to the next region on the left where x < 0, expression will be positive again because now both factors x and (x - 3) are negative. The product of two negatives is positive so the expression will be positive.
Similarly, we can solve a question with any number of factors.
Show more


karishma great that was very helpful one last question what if inequality was
x(x - 3) > 0 instead of
x(x - 3) < 0

Will it still hold true
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puneetfitness



karishma great that was very helpful one last question what if inequality was
x(x - 3) > 0 instead of
x(x - 3) < 0

Will it still hold true
Show more

Then look for the regions with the positive sign on your number line.

The solution will be x < 0 or x > 3
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If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x 01 Jul 2023, 18:27
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boomtangboy
If x ≠ 0, is |x| < 1 ?
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Whether this can be deduced\(-1<x<1\)

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

\(-1<x< 1\)

Answer to the Question is YES, hence, sufficient

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

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

This is only possible for

\(0<x< 1\)

Answer to the Question is YES, hence, sufficient

Therefore, Opt (D)
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If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x 13 Aug 2023, 22:15
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shikhar
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
Show more
Shikhar you have to check two sides of the absolute value. If you solve that equation in case
x>0 solution will be (0,1) interval
second case
x<0 solution is (-1,0) interval
combo is what we need
(-1,0) v (0,1)

Posted from my mobile device
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If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x 28 Oct 2023, 03:10
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KarishmaB
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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.
Show more


Hi KarishmaB great explanation.
One question here
(x^2 - 1)/x < 0
x^2 < 1
x<+-1
x< 1 or x < -1

Therefore not sure how is this become -1< x < 1 ?
Could you help clarify? Thanks
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If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x 28 Oct 2023, 12:02
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Bunuel
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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.
Show more

Hi Bunuel, LHS (|x|) is an absolute value therefore x could be both positive and negative. So not sure why you mean |x| is always non-negative here?
RHS (1/x), if x is -2, then |x| < 1/x mean -2 < -1/2. Therefore x could be negative here.
Not sure where's my thinking going wrong? Could you help clarify? Thanks
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Kimberly77
KarishmaB
Shef08
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.


Hi KarishmaB great explanation.
One question here
(x^2 - 1)/x < 0
x^2 < 1
x<+-1
x< 1 or x < -1

Therefore not sure how is this become -1< x < 1 ?
Could you help clarify? Thanks
Show more

When we have inequalities with linear factors, we plot them on the number line (search for wavy curve method) to get our range of interest.

(x^2 - 1)/x = (x + 1)(x - 1)/x

So plot -1, 0 and 1 on the number line. Since we want the negative regions, 0 < x < 1 and x < -1
Since x must be positive, the acceptable range is only 0 < x < 1

I will put up a YouTube video on this concept soon.
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If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x 02 Nov 2023, 12:18
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Thanks @KarishmaB...still not quite sure about the negative regions and and x must be positive here...
Looking forward to your YouTube video and thanks a bunch :please: :)
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