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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/(x2) >= 4, b)x3<2x4 c)x + 2 < x2  4 solve for x I will highly appreciate detailed explanations, thanks



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A. 1/(x2) >= 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:
x2 =< 1/4
That means,
o if x > 2 then,
x2 =< 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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B. x3<2x4
o If x < 3, then the inequation becomes:
(x3)<2x4
<=> 7 < 3x
<=> x > 7/3
=> 7/3 < x < 3
o If x >= 3, then the inequation becomes:
x3<2x4
<=> 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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C. x + 2 < x2  4
x2  4 = (x+2)(x2)
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)(x2)) > 0 when x > 2 or x < 2
o Sign((x+2)(x2)) < 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)(x3) > 0
One more time, a is positive here. Thus (x+2)(x3) > 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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FIG I HAVE A QUESTION HERE ABOUT THE ANSWER
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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yezz wrote: FIG I HAVE A QUESTION HERE ABOUT THE ANSWER
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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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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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 24 = 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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Thanks man , this is no catch this the fruits of your help and coaching
thanks a lot



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Fig wrote: B. x3<2x4
o If x < 3, then the inequation becomes: (x3)<2x4 <=> 7 < 3x <x> 7/3 => 7/3 < x <3>= 3, then the inequation becomes:[/b] x3<2x4 <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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ajay_gmat wrote: Fig wrote: B. x3<2x4
o If x < 3, then the inequation becomes: (x3)<2x4 <=> 7 < 3x <x> 7/3 => 7/3 < x <3>= 3, then the inequation becomes:[/b] x3<2x4 <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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Joined: 30 Jul 2009
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Fig wrote: A. 1/(x2) >= 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:
x2 =< 1/4
can someone please explain why this happens. Why can't I just use 1/x2 >=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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Fig wrote: B. x3<2x4
o If x < 3, then the inequation becomes: (x3)<2x4 <=> 7 < 3x <=> x > 7/3 => 7/3 < x < 3
o If x >= 3, then the inequation becomes: x3<2x4 <=> x > 1 => x >= 3
Finally, we have : x > 7/3 Hi buddy could you please explain this, <=> 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



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Fig wrote: C. x + 2 < x2  4
x2  4 = (x+2)(x2)
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)(x2)) > 0 when x > 2 or x < 2 o Sign((x+2)(x2)) < 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)(x2) Solving for x in x^2  4 > x + 2 in the following two scenarios: 1. When (x+2)(x2) >= 0 ==> x^2  4 > x + 2 ==> x>1 and x>2 2. When (x+2)(x2) < 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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fortsill wrote: Fig wrote: C. x + 2 < x2  4
x2  4 = (x+2)(x2)
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)(x2)) > 0 when x > 2 or x < 2 o Sign((x+2)(x2)) < 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)(x2) Solving for x in x^2  4 > x + 2 in the following two scenarios: 1. When (x+2)(x2) >= 0 ==> x^2  4 > x + 2 ==> x>1 and x>2 2. When (x+2)(x2) < 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: \(x^24>x+2\). We should consider the cases when \(x^24<0\) and \(x^24\geq{0}\): \(x^24<0\) for \(2<x<2\) > \((x^24)>x+2\) > \(x^2+x2<0\) > \((x+2)(x1)<0\)> \(2<x<1\). Since we are considering the range \(2<x<2\) then the solution for this case will be intersections of these two (common part): \(2<x<1\); \(x^24\geq{0}\) for \(x\leq{2}\) or \(x\geq{2}\) > \((x^24)>x+2\) > \(x^2x6>0\) > \((x+2)(x3)>0\)> \(x<2\) 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 \(x^24>x+2\) true are: \(x<2\), \(2<x<1\) and \(x>3\). Solving inequalities: x24x94661.html#p731476inequalitiestrick91482.htmldatasuffinequalities109078.htmlrangeforvariablexinagiveninequality109468.html?hilit=extreme#p873535everythingislessthanzero108884.html?hilit=extreme#p868863Hope it helps.
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Re: For everyone interested in modulus practices, I have listed, [#permalink]
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Re: For everyone interested in modulus practices, I have listed,
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