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# For everyone interested in modulus practices, I have listed,

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For everyone interested in modulus practices, I have listed, [#permalink]

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03 Dec 2006, 05:22
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For everyone interested in modulus practices, I have listed, at the bottom of this topic, several threads of modulus in a graduated level of difficulty.

Keep in mind that it's for trainning purposes and that the most advanced posts are not reprensative of the real GMAT.

Finally, enjoy !..... Fig

a) |1/(x-2)| >= 4,
b)|x-3|<2x-4
c)x + 2 < |x2 - 4|

solve for x

I will highly appreciate detailed explanations, thanks
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03 Dec 2006, 05:45
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A. |1/(x-2)| >= 4

As we are working positive value and as the function f(x) = 1/x decreases when x increases on the domain of x > 0, we can imply that:

|x-2| =< 1/4

That means,

o if x > 2 then,
x-2 =< 1/4
<=> x <= 1/4+2
<=> x <= 9/4

o if x < 2 then,
-x+2 =< 1/4
<=> x >= 2 - 1/4
<=> x >= 7/4

Thus, 7/4 <= x <= 9/4

Last edited by Fig on 29 Dec 2006, 13:57, edited 2 times in total.
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04 Dec 2006, 10:39
B. |x-3|<2x-4

o If x < 3, then the inequation becomes:
-(x-3)<2x-4
<=> 7 < 3x
<=> x > 7/3
=> 7/3 < x < 3

o If x >= 3, then the inequation becomes:
x-3<2x-4
<=> x > 1
=> x >= 3

Finally, we have : x > 7/3

Last edited by Fig on 04 Dec 2006, 13:40, edited 1 time in total.
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04 Dec 2006, 13:37
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C. x + 2 < |x2 - 4|

|x2 - 4| = |(x+2)(x-2)|

Since the sign of a*x^2+b*x+c is that of a for any values of x not between the 2 roots, solutions of the equation a*x^2+b*x+c = 0 if they exist.

Thus,
o Sign((x+2)(x-2)) > 0 when x > 2 or x < -2
o Sign((x+2)(x-2)) < 0 when -2 < x < 2

If -2 < x < 2, the inequation becomes :
x + 2 < -(x^2 - 4)
<=> x^2 + x - 2 < 0 (B)

Delta = 1 + 4*2 = 9.

The roots are:
o Root1 = (-1 - 3)/2 = -2
o Root2 = (-1 + 3)/2 = 1

So, (B) is true when x is between the roots of x^2 + x - 2. That implies -2 < x < 1.

If x < -2 or x > 2, the inequation becomes :
x + 2 < x^2 - 4
<=> x^2 - x - 6 > 0 (A)

Delta = 1 + 4*6 = 25.

The roots are:
o Root1 = (1 - 5)/2 = -2
o Root2 = (1 + 5)/2 = 3

(A) : x^2 - x - 6 > 0
<=> (x+2)(x-3) > 0

One more time, a is positive here. Thus (x+2)(x-3) > 0 outside of the roots.
In other words, x > 3 or x < -2

Finally, the domain of solutions of x is : x > 3 or x < -2 or -2 < x < 1.

Last edited by Fig on 29 Dec 2006, 04:26, edited 2 times in total.
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05 Dec 2006, 02:51
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THE RANGE , 7/4 <= x <= 9/4 INCLUDES 2 , AND THE FUNCTION IS NOT DEFINED AT X=2 AM I RIGHT???
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05 Dec 2006, 04:20
yezz wrote:

THE RANGE , 7/4 <= x <= 9/4 INCLUDES 2 , AND THE FUNCTION IS NOT DEFINED AT X=2 AM I RIGHT???

Yes .... U are right... My conclusion should clarify it .... We have to remove the x=2
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10 Dec 2006, 05:22
Hello Fig , hope u r fine and in best health ,

FOR THE THIRD PROBLEM , I BELIEVE THE ANSWER MISS THE RANGE

-2<X<1
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10 Dec 2006, 10:06
yezz wrote:
Hello Fig , hope u r fine and in best health ,

FOR THE THIRD PROBLEM , I BELIEVE THE ANSWER MISS THE RANGE

-2<X<1

I'm ok thanks... it follows the way

Great catch .... Yes, I did a "silly" mistake 2-4 = 6 and ended up with a wrong x^2 + x + 6 .... I have corrected my post following your catch... very good :D
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13 Dec 2006, 06:07
Thanks man , this is no catch this the fruits of your help and coaching

thanks a lot
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28 Dec 2006, 23:54
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29 Jul 2007, 06:56
Fig wrote:
B. |x-3|<2x-4

o If x < 3, then the inequation becomes:
-(x-3)<2x-4
<=> 7 < 3x
<x> 7/3
=> 7/3 < x <3>= 3, then the inequation becomes:[/b]
x-3<2x-4
<x> 1
=> x >= 3

Finally, we have : x > 7/3

Hi Fig,

Why have you mentioned that x >=3 and not that x > 3.. I see absolute Questions which sometimes have the >= sign and sometimes the > sign only.. Might be a very mindless question but hope you would answer..
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29 Jul 2007, 13:48
ajay_gmat wrote:
Fig wrote:
B. |x-3|<2x-4

o If x < 3, then the inequation becomes:
-(x-3)<2x-4
<=> 7 < 3x
<x> 7/3
=> 7/3 < x <3>= 3, then the inequation becomes:[/b]
x-3<2x-4
<x> 1
=> x >= 3

Finally, we have : x > 7/3

Hi Fig,

Why have you mentioned that x >=3 and not that x > 3.. I see absolute Questions which sometimes have the >= sign and sometimes the > sign only.. Might be a very mindless question but hope you would answer..

Not meaningless but rather precised question ...

Actually, it's just to avoid adding 1 solution alone from another interval... and so to meet a higher degree of "perfection" by making it a smoother way to the final solution and a straightforward simplicity to understand

Note that nothing is wrong if we say > or >= as soon as we do not miss to consider all values of x, especially here the ones that limit 2 intervals
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09 Sep 2009, 18:01
Fig wrote:
A. |1/(x-2)| >= 4

As we are working positive value and as the function f(x) = 1/x decreases when x increases on the domain of x > 0, we can imply that:

|x-2| =< 1/4

can someone please explain why this happens. Why can't I just use |1/x-2| >=4. However, when x<2, I get x<=7/4. Also I agree with everyone else as the range include 2, doesn't lie outside of our limitations.
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15 Dec 2009, 19:31
Fig wrote:
B. |x-3|<2x-4

o If x < 3, then the inequation becomes:
-(x-3)<2x-4
<=> 7 < 3x
<=> x > 7/3
=> 7/3 < x < 3

o If x >= 3, then the inequation becomes:
x-3<2x-4
<=> x > 1
=> x >= 3

Finally, we have : x > 7/3

Hi buddy

<=> x > 7/3
=> 7/3 < x < 3
<=> x > 1
=> x >= 3

i solved and got x>1 & x>7/3
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Re: Absolute Mpractices [Modules] - List of GMAT Questions [#permalink]

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12 Aug 2010, 05:10
in case If -2 < x < 2, shouldn't it also have one more solution to problem.i.e. x<-2 and x>1
Intern
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24 Feb 2012, 23:27
Fig wrote:
C. x + 2 < |x2 - 4|

|x2 - 4| = |(x+2)(x-2)|

Since the sign of a*x^2+b*x+c is that of a for any values of x not between the 2 roots, solutions of the equation a*x^2+b*x+c = 0 if they exist.

Thus,
o Sign((x+2)(x-2)) > 0 when x > 2 or x < -2
o Sign((x+2)(x-2)) < 0 when -2 < x < 2

Trying hard to grasp this concept, but don't think I'm understanding what you're trying to convey here.
Ideally, I would try to solve this as such:

Given that |x^2 - 4| is same as |(x+2)(x-2)|

Solving for x in |x^2 - 4| > x + 2 in the following two scenarios:
1. When (x+2)(x-2) >= 0
==> x^2 - 4 > x + 2
==> x>-1 and x>2

2. When (x+2)(x-2) < 0
==> -(x^2 - 4) > x + 2
==> 4 - x^2 > x + 2
==> x<1 and x<-2

Does the combination of these solutions for x represent the final answer? Am I missing a step?
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25 Feb 2012, 02:08
Expert's post
fortsill wrote:
Fig wrote:
C. x + 2 < |x2 - 4|

|x2 - 4| = |(x+2)(x-2)|

Since the sign of a*x^2+b*x+c is that of a for any values of x not between the 2 roots, solutions of the equation a*x^2+b*x+c = 0 if they exist.

Thus,
o Sign((x+2)(x-2)) > 0 when x > 2 or x < -2
o Sign((x+2)(x-2)) < 0 when -2 < x < 2

Trying hard to grasp this concept, but don't think I'm understanding what you're trying to convey here.
Ideally, I would try to solve this as such:

Given that |x^2 - 4| is same as |(x+2)(x-2)|

Solving for x in |x^2 - 4| > x + 2 in the following two scenarios:
1. When (x+2)(x-2) >= 0
==> x^2 - 4 > x + 2
==> x>-1 and x>2

2. When (x+2)(x-2) < 0
==> -(x^2 - 4) > x + 2
==> 4 - x^2 > x + 2
==> x<1 and x<-2

Does the combination of these solutions for x represent the final answer? Am I missing a step?

First of all you won't see such equation on the GMAT, but anyway:

$$|x^2-4|>x+2$$.

We should consider the cases when $$x^2-4<0$$ and $$x^2-4\geq{0}$$:

$$x^2-4<0$$ for $$-2<x<2$$ --> $$-(x^2-4)>x+2$$ --> $$x^2+x-2<0$$ --> $$(x+2)(x-1)<0$$--> $$-2<x<1$$. Since we are considering the range $$-2<x<2$$ then the solution for this case will be intersections of these two (common part): $$-2<x<1$$;

$$x^2-4\geq{0}$$ for $$x\leq{-2}$$ or $$x\geq{2}$$ --> $$(x^2-4)>x+2$$ --> $$x^2-x-6>0$$ --> $$(x+2)(x-3)>0$$--> $$x<-2$$ or $$x>3$$. Since we are considering the range $$x\leq{-2}$$ or $$x\geq{2}$$ then the solution for this case will be intersections of these four (common parts): $$x<-2$$ or $$x>3$$;

Finally the ranges of $$x$$ for which $$|x^2-4|>x+2$$ true are: $$x<-2$$, $$-2<x<1$$ and $$x>3$$.

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.
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Re: For everyone interested in modulus practices, I have listed, [#permalink]

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27 Aug 2014, 04:40
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Re: For everyone interested in modulus practices, I have listed, [#permalink]

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02 Feb 2016, 23:21
Hello from the GMAT Club BumpBot!

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Re: For everyone interested in modulus practices, I have listed,   [#permalink] 02 Feb 2016, 23:21
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