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Which of the following represents the complete range of x [#permalink]
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subhashghosh wrote:
Hi Bunuel

I'm getting D as answer :

x^3(1-2x)(1+2x) < 0

\(-ve --- -1/2---- +ve--- 0----- -ve-----1/2--- +ve\)
Could you please explain where I'm wrong ?

Regards,
Subhash


Even though your question is directed to Bunuel, I will give a quick explanation.

The concept of the rightmost section being positive is applicable when every term is positive in the rightmost region. This is the case whenever the expressions involved are of the form (x - a) or (ax - b) etc. When you have a term such as (1-2x), the rightmost region becomes negative. So either, as Bunuel mentioned, check for an extreme value of x or convert (1-2x) to (2x - 1) and flip the sign to >.

Check out this post for more on complete range questions: https://anaprep.com/algebra-must-be-tru ... questions/

Originally posted by KarishmaB on 13 Feb 2011, 22:05.
Last edited by KarishmaB on 24 Aug 2023, 06:10, edited 2 times in total.
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Re: Which of the following represents the complete range of x [#permalink]
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Thanx a trillion for your post on solving inequalities using graph
You know i paid over 300$ to test prep institutes but got nothing out of it.......when i asked such basic question the tutor got frustrated and insulted me.....But hats off to you... MAx wat will i give 1 kudo......
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Re: Which of the following represents the complete range of x [#permalink]
Thanks Bunuel. +1

A question - what is the best way u use to know if the "good" area is above or below?

i mean - what was the best way for u to know that its between -1/2 to 0

i used numbers ex. 1/4 but it consumes time! is there any better technique?

thanks.
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Re: Which of the following represents the complete range of x [#permalink]
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Hi Bunuel

I'm getting D as answer :

x^3(1-2x)(1+2x) < 0

\(-ve --- -1/2---- +ve--- 0----- -ve-----1/2--- +ve\)
Could you please explain where I'm wrong ?

Regards,
Subhash
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Re: Which of the following represents the complete range of x [#permalink]
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Karishma
I flipped the sign before. So I got x^3(2x-1)(2x-1) > 0

2 cases - both +ve or both -ve

case 1
-------
x > 0 and |x| > 1/2. Hence x > 1/2

case 2
------
x < 0 and 4x^2 - 1 < 0
x < 0 and -1/2 < x < 1/2
Taking the most restrictive value-
-1/2 < x < 0

I hope this is correct. Btw this is 750 level in 2 mins.

VeritasPrepKarishma wrote:
subhashghosh wrote:
Hi Bunuel

I'm getting D as answer :

x^3(1-2x)(1+2x) < 0

\(-ve --- -1/2---- +ve--- 0----- -ve-----1/2--- +ve\)
Could you please explain where I'm wrong ?

Regards,
Subhash


Even though your question is directed to Bunuel, I will give a quick explanation.

The concept of the rightmost section being positive is applicable when every term is positive in the rightmost region. This is the case whenever the expressions involved are of the form (x - a) or (ax - b) etc. When you have a term such as (1-2x), the rightmost region becomes negative. So either, as Bunuel mentioned, check for an extreme value of x or convert (1-2x) to (2x - 1) and flip the sign to >.
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gmat1220 wrote:
Karishma
I flipped the sign before. So I got x^3(2x-1)(2x-1) > 0

2 cases - both +ve or both -ve

case 1
-------
x > 0 and |x| > 1/2. Hence x > 1/2

case 2
------
x < 0 and 4x^2 - 1 < 0
x < 0 and -1/2 < x < 1/2
Taking the most restrictive value-
-1/2 < x < 0

I hope this is correct. Btw this is 750 level in 2 mins.


Yes, it is correct... and since you know what you are doing, you will need to work very hard to fall short of time on GMAT.
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gmatpapa wrote:
Which of the following represents the complete range of x over which x^3 - 4x^5 < 0?

(A) 0 < |x| < ½
(B) |x| > ½
(C) –½ < x < 0 or ½ < x
(D) x < –½ or 0 < x < ½
(E) x < –½ or x > 0


Responding to a pm:
The problem is the same here.
How do you solve this inequality: \((1+2x)*x^3*(1-2x)<0\)

Again, there are 2 ways -
The long algebraic method: When is \((1+2x)*x^3*(1-2x)\) negative? When only one of the terms is negative or all 3 are negative. There will be too many cases to consider so this is painful.

The number line method: Multiply both sides of \((1+2x)*x^3*(1-2x)<0\) by -1 to get \((2x + 1)*x^3*(2x - 1)>0\)
Take out 2 common to get \(2(x + 1/2)*x^3*2(x - 1/2)>0\) [because you want each term to be of the form (x + a) or (x - a)]
Now plot them on the number line and get the regions where this inequality holds.

Originally posted by KarishmaB on 20 Jun 2012, 22:42.
Last edited by KarishmaB on 23 Aug 2023, 00:40, edited 1 time in total.
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Re: Which of the following represents the complete range of x [#permalink]
Hi All,

Could I conclude that for this case i.e (1+2x)*x^3*(1-2x)<0
even if one of the terms <0, that does not necessarily mean that the entire product of the 3 terms <0.
Cause like if the eq was (1+2x)*x^3*(1-2x)= 0 ....I could have safely concluded that
However in this case for the entire product <0.. either 1 terms or 2 terms or even all 3 terms can be - ve.
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lesnin wrote:
Hi All,

Could I conclude that for this case i.e (1+2x)*x^3*(1-2x)<0
even if one of the terms <0, that does not necessarily mean that the entire product of the 3 terms <0.
Cause like if the eq was (1+2x)*x^3*(1-2x)= 0 ....I could have safely concluded that
However in this case for the entire product <0.. either 1 terms or 2 terms or even all 3 terms can be - ve.


When you have product of two or more terms, the product will be negative only when odd number of terms are negative i.e. either only one term is negative and rest are positive or only 3 terms are negative and rest are positive or only 5 terms are negative and rest are positive.
(-)(+)(+) = (-)
(-)(-)(+) = (+)
(-)(-)(-) = (-)
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Using the amazing technique:

\(x^3(1-4x^2)<0\)
\(x^3(1-2x)(1+2x)<0\)

+ (-1/2) - (0) + (1/2) -

If less than 0, select (-) curves.

Answer: -1/2 < x < 0 or 1/2 < x ==> C
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Re: Which of the following represents the complete range of x [#permalink]
This might be a difficult question to answer, but here it is:

I understand the methodology in how the correct answer was arrived at (thanks, Bunuel) but my question is, how do I know to use that methodology with this particular question?

Also, could I solve for this problem using x^3(1-4x^2)<0 as opposed to (1+2x)*x^3*(1-2x)<0?

As always, thanks to the community for all of your help.
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Re: Which of the following represents the complete range of x [#permalink]
Which of the following represents the complete range of x over which x^3 – 4x^5 < 0?

x^3 – 4x^5 < 0
x^3(1-4x^2) < 0
(1-4x^2) < 0
1 < 4x^2
√1 < √4x^2
(when you take the square root of 4x^2 you take the square root of a square so...)
1 < |2x|

1<(2x)
1/2 < x
OR
1<-2x
-1/2>x

I am still a bit confused as to how we get 0. I see how it is done with the "root" method but my way of solving was just a bit different. Any thoughts?
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WholeLottaLove wrote:
Which of the following represents the complete range of x over which x^3 – 4x^5 < 0?

x^3 – 4x^5 < 0
x^3(1-4x^2) < 0
(1-4x^2) < 0
1 < 4x^2
√1 < √4x^2
(when you take the square root of 4x^2 you take the square root of a square so...)
1 < |2x|

1<(2x)
1/2 < x
OR
1<-2x
-1/2>x

I am still a bit confused as to how we get 0. I see how it is done with the "root" method but my way of solving was just a bit different. Any thoughts?


The step in red above is your problem. How did you get rid of x^3? Can you divide both sides by x^3 when you have an inequality? You don't know whether x^3 is positive or negative. If you divide both sides by x^3 and x^3 is negative, the sign will flip. So you must retain the x^3 and that will give you 3 transition points (-1/2, 0 , 1/2)
Even in equations, it is not a good idea to cancel off x from both sides. You might lose a solution in that case x = 0
e.g.
x(x - 1) = 0
(x - 1) = 0
x = 1 (Incomplete)


x(x-1) = 0
x = 0 or 1 (Correct)
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WholeLottaLove wrote:
This might be a difficult question to answer, but here it is:

I understand the methodology in how the correct answer was arrived at (thanks, Bunuel) but my question is, how do I know to use that methodology with this particular question?

Also, could I solve for this problem using x^3(1-4x^2)<0 as opposed to (1+2x)*x^3*(1-2x)<0?

As always, thanks to the community for all of your help.


When you have linear factors and inequalities, think of this method. Since this method is useful for linear factors, you need to split the quadratic (1 - 4x^2) into (1-2x)*(1+2x).
Some quadratic or higher powers may not be a problem (e.g. (x + 1)^2, (x^2 + 1) etc are always positive) so they can be ignored.

Originally posted by KarishmaB on 01 Aug 2013, 23:29.
Last edited by KarishmaB on 23 Aug 2023, 00:42, edited 1 time in total.
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Re: Which of the following represents the complete range of x [#permalink]
Bunuel wrote:
gmatpapa wrote:
Which of the following represents the complete range of x over which x^3 - 4x^5 < 0?

(A) 0 < |x| < ½
(B) |x| > ½
(C) –½ < x < 0 or ½ < x
(D) x < –½ or 0 < x < ½
(E) x < –½ or x > 0


Basically we are asked to find the range of \(x\) for which \(x^3-4x^5<0\) is true.

\(x^3-4x^5<0\) --> \(x^3(1-4x^2)<0\) --> \((1+2x)*x^3*(1-2x)<0\) --> roots are -1/2, 0, and 1/2 --> \(-\frac{1}{2}<x<0\) or \(x>\frac{1}{2}\).

Answer: C.




Check this for more: inequalities-trick-91482.html




Hi Bunuel,
I tried the trick, however using the equation I am getting different ranges.
below is what I did ..
1) f(x) <0
2) roots are -1/2 , 0, 1/2

- (-1/2) + 0 - 1/2 +
starting from + from right.

now as per this
x< -1/2 and 0<x<1/2

can you advice where I went wrong...
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seabhi wrote:
Bunuel wrote:
gmatpapa wrote:
Which of the following represents the complete range of x over which x^3 - 4x^5 < 0?

(A) 0 < |x| < ½
(B) |x| > ½
(C) –½ < x < 0 or ½ < x
(D) x < –½ or 0 < x < ½
(E) x < –½ or x > 0


Basically we are asked to find the range of \(x\) for which \(x^3-4x^5<0\) is true.

\(x^3-4x^5<0\) --> \(x^3(1-4x^2)<0\) --> \((1+2x)*x^3*(1-2x)<0\) --> roots are -1/2, 0, and 1/2 --> \(-\frac{1}{2}<x<0\) or \(x>\frac{1}{2}\).

Answer: C.




Check this for more: inequalities-trick-91482.html




Hi Bunuel,
I tried the trick, however using the equation I am getting different ranges.
below is what I did ..
1) f(x) <0
2) roots are -1/2 , 0, 1/2

- (-1/2) + 0 - 1/2 +
starting from + from right.

now as per this
x< -1/2 and 0<x<1/2

can you advice where I went wrong...


The factors must be of the form (x - a), (x - b) etc. Notice that one factor here is of the form (1 - 2x). You need to change this.

\((1+2x)*x^3*(1-2x)<0\)
\(2(x + 1/2)*x^3*2(x - 1/2) > 0\) (note the sign flip)

Now the factors are of the form required and it is clear that the transition points are -1/2, 0, 1/2.

The required range is x > 1/2 or -1/2 < x< 0
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