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Re: not that hard [#permalink]
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07 Sep 2010, 05:31
Oh, yes, you are right, Bunuel. Thanks.



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Re: not that hard [#permalink]
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16 Feb 2012, 23:02
Hi, If i solve option 1. i get x^21>0 or (x+1)(x1)>0 this implies : x>1 or x>1.. however the question in the stem can be rephrased as x<1 or X>1.. How is option 1 true then?am i missing something? Bunuel wrote: mehdiov wrote: If x<x^2 , which of the following must be true?
I. x^2>1 II. x>0 III. x<1
(A) I only (B) II only (C) III only (D) I and II only (E) I and III only
what is the fastest way to solbe it ? Given: \(x<x^2\) > reduce by \(x\) (side note: we can safely do this as absolute value is nonnegative and in this case we know it's not zero too, as if x would be zero inequality wouldn't hold true, so \(x>0\)) > \(1<x\) > \(x<1\) or \(x>1\). So we have that \(x<1\) or \(x>1\). I. x^2>1 > always true; II. x>0 > may or may not be true; III. x<1 > > may or may not be true. Answer: A (I only).



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Re: not that hard [#permalink]
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16 Feb 2012, 23:06



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Re: not that hard [#permalink]
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21 May 2012, 08:28
Bunuel wrote: mehdiov wrote: If x<x^2 , which of the following must be true?
I. x^2>1 II. x>0 III. x<1
(A) I only (B) II only (C) III only (D) I and II only (E) I and III only
what is the fastest way to solbe it ? Given: \(x<x^2\) > reduce by \(x\) (side note: we can safely do this as absolute value is nonnegative and in this case we know it's not zero too, as if x would be zero inequality wouldn't hold true, so \(x>0\)) > \(1<x\) > \(x<1\) or \(x>1\). So we have that \(x<1\) or \(x>1\). I. x^2>1 > always true; II. x>0 > may or may not be true; III. x<1 > > may or may not be true. Answer: A (I only). Bunuel x.x=x^2 so when we divide x^2 by x we get x so x.x= x^2 is this a formula



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Re: not that hard [#permalink]
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21 May 2012, 09:00
Joy111 wrote: Bunuel wrote: mehdiov wrote: If x<x^2 , which of the following must be true?
I. x^2>1 II. x>0 III. x<1
(A) I only (B) II only (C) III only (D) I and II only (E) I and III only
what is the fastest way to solbe it ? Given: \(x<x^2\) > reduce by \(x\) (side note: we can safely do this as absolute value is nonnegative and in this case we know it's not zero too, as if x would be zero inequality wouldn't hold true, so \(x>0\)) > \(1<x\) > \(x<1\) or \(x>1\). So we have that \(x<1\) or \(x>1\). I. x^2>1 > always true; II. x>0 > may or may not be true; III. x<1 > > may or may not be true. Answer: A (I only). Bunuel x.x=x^2 so when we divide x^2 by x we get x so x.x= x^2 is this a formula Yes, x*x=x^2. Consider x=2 then x*x=2*2=4=(2)^2.
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Re: If x<x^2 , which of the following must be true? [#permalink]
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23 May 2012, 05:10
Hi Bunuel, I understood your logic that we can safely divide by x, however, I tried to solve it this way, just wanted to confirm is this is also right...
x2>x => if x is positive then x2x>0 => x(x1)>0 => x<0 and x>1................case 1
and if x is negative, then x2+x>0 => x(x+1)>0 =>x<1 and x>0...case2
From case 1 and case2....x<1 and x>1..[Same conclusion but lengthier method].. I want to especially confirm because I tried to apply modulus and inequalities theory.. so want to be sure.



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Re: If x<x^2 , which of the following must be true? [#permalink]
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23 May 2012, 12:09
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picked A in 46 secs. for X to be less than x^2 just means that it is x is not a decimal. it can be less than 1 and greater than 1. you can only say this in totality with certainty. And square of nay no less than 1 and greater than 1 will be greater than 1. so A



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Re: If x<x^2 , which of the following must be true? [#permalink]
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23 May 2012, 12:54
Bunuel: I have solved the problem in the following manner: Absolute value of x is positive. So, we can divide both side by x. that leaves us, 1<x or x>1. or x^2>1 Though i have the OA, am i making any fundamental error?



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Re: If x<x^2 , which of the following must be true? [#permalink]
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24 May 2012, 00:45



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Re: If x<x^2 , which of the following must be true? [#permalink]
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24 May 2012, 18:09
Bunuel, Doesn't the question mean  which of the values of x must hold for X<X^2 all values of x<1 satisfy , although there are other solutions(x>1) but E does not say those solution do not exist. In fact if a superset of values are true then subset of values must be true...
where am misinterpreting



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Re: If x<x^2 , which of the following must be true? [#permalink]
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24 May 2012, 20:48
mehdiov wrote: If x<x^2 , which of the following must be true?
I. x^2>1 II. x>0 III. x<1
(A) I only (B) II only (C) III only (D) I and II only (E) I and III only
what is the fastest way to solbe it ? Hi, Using number lines: for x > 0, \(x < x^2\) x(x+1) > 0, number line (a) for x < 0,\(x < x^2\) x(x1) > 0, number line (b) (a) (1)0 (b)01 combining both we have, x>1, x<1 \(x^2 > 1\), already includes above the above values of x. Thus, answer is (A).



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Re: If x<x^2 , which of the following must be true? [#permalink]
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24 May 2012, 21:16
vikram 4689 what x <1 says is that x can ONLY hold values of less than 1. Which actually means that x cannot have values of 2,3.. etc. this we all know is not true as x can also takes values of >1. Hence this statement cannot hold true. makes sense?



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Re: If x<x^2 , which of the following must be true? [#permalink]
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25 May 2012, 02:01
vikram4689 wrote: Bunuel, Doesn't the question mean  which of the values of x must hold for X<X^2 all values of x<1 satisfy , although there are other solutions(x>1) but E does not say those solution do not exist. In fact if a superset of values are true then subset of values must be true...
where am misinterpreting No, that's not true. Again: \(x<x^2\) is given as a fact and then we asked to determine which of the following statements MUST be true. \(x<x^2\) means that \(x<1\) or \(x>1\). Now, evaluate each option: I. x^2>1. Since \(x<1\) or \(x>1\) then this option is always true; II. x>0. Since \(x<1\) or \(x>1\) then this option may or may not be true; III. x<1. Since \(x<1\) or \(x>1\) then this option may or may not be true. Answer: A (I only).
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Re: If x<x^2 , which of the following must be true? [#permalink]
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25 May 2012, 02:22
If x<x^2 , which of the following must be true? I. x^2>1 II. x>0 III. x<1 (A) I only (B) II only (C) III only (D) I and II only (E) I and III only
what is the fastest way to solbe it ? hope this might help u .. x^2 >x> (implies) x^2>x (since x^2 can not be less than ..it will always by +ve) x^2x>0 x(x1)>0 this would give two values x <0 and x>1 but x can not be <0 because values lying between 1 to +1 would nt satisfy the given condition x<x^2 ...therefore... x>1 therefore x^2 is also greater than 1



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Re: If x<x^2 , which of the following must be true? [#permalink]
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25 May 2012, 02:25



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Re: If x<x^2 , which of the following must be true? [#permalink]
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25 May 2012, 02:36
Sorry i did a mishtake...i would like to revise it,.....
If x<x^2 , which of the following must be true? I. x^2>1 II. x>0 III. x<1 (A) I only (B) II only (C) III only (D) I and II only (E) I and III only
what is the fastest way to solbe it ? hope this might help u .. x^2 >x> (implies) i)x^2>x x^2x>0 x(x1)>0 this would give two values x <0 and x>1
ii) x^2 <x x^2x<0 x(x+1)<0 giving u ..x>0 and x <1
giving u the common condition that is x >1 therefore x^2 is also greater than 1



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Re: If x<x^2 , which of the following must be true? [#permalink]
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25 May 2012, 02:40



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Re: If x<x^2 , which of the following must be true? [#permalink]
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25 May 2012, 21:41
Bunuel wrote: pavanpuneet wrote: Hi Bunuel, I understood your logic that we can safely divide by x, however, I tried to solve it this way, just wanted to confirm is this is also right...
x2>x => if x is positive then x2x>0 => x(x1)>0 => x<0 and x>1................case 1
and if x is negative, then x2+x>0 => x(x+1)>0 =>x<1 and x>0...case2
From case 1 and case2....x<1 and x>1..[Same conclusion but lengthier method].. I want to especially confirm because I tried to apply modulus and inequalities theory.. so want to be sure. x^2>x If x>0 then x2x>0 > x(x1)>0 > x<0 and x>1. Since we are considering x>0 range then x>1; If x<0 then x2x>0 > x(x+1)>0 > x<1 and x>0. Since we are considering x<0 range then x<1; So, x^2>x holds true for x<1 and x>1. Hope it's clear. Hello Bunuel I didnot understand this part x^2>x holds true for x<1 and x>1. if you have x<1 then why ans "E" is not true? i also didnot understand folloiwng part "If x>0 then x2x>0 > x(x1)>0 >[highlight]x<0[/highlight] and x>1" how can x<0 ? I solved this problem and voted for "A" for x>1 / 1>x>0 / x<0 (I considered these 3 ranged because if 1>x>0 then the square of number is less then number) if (x>1) > x^2>x holds true if (1>x>0) > x^2>x doesnot holds true if (x<0) > x^2>x holds true (1/2)^2 > (1/2)



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Re: If x<x^2 , which of the following must be true? [#permalink]
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26 May 2012, 02:36
kuttingchai wrote: Bunuel wrote: pavanpuneet wrote: Hi Bunuel, I understood your logic that we can safely divide by x, however, I tried to solve it this way, just wanted to confirm is this is also right...
x2>x => if x is positive then x2x>0 => x(x1)>0 => x<0 and x>1................case 1
and if x is negative, then x2+x>0 => x(x+1)>0 =>x<1 and x>0...case2
From case 1 and case2....x<1 and x>1..[Same conclusion but lengthier method].. I want to especially confirm because I tried to apply modulus and inequalities theory.. so want to be sure. x^2>x If x>0 then x2x>0 > x(x1)>0 > x<0 and x>1. Since we are considering x>0 range then x>1; If x<0 then x2x>0 > x(x+1)>0 > x<1 and x>0. Since we are considering x<0 range then x<1; So, x^2>x holds true for x<1 and x>1. Hope it's clear. Hello Bunuel I didnot understand this part x^2>x holds true for x<1 and x>1. if you have x<1 then why ans "E" is not true? i also didnot understand folloiwng part "If x>0 then x2x>0 > x(x1)>0 >[highlight]x<0[/highlight] and x>1" how can x<0 ? E cannot be correct since I solved this problem and voted for "A" for x>1 / 1>x>0 / x<0 (I considered these 3 ranged because if 1>x>0 then the square of number is less then number) if (x>1) > x^2>x holds true if (1>x>0) > x^2>x doesnot holds true if (x<0) > x^2>x holds true (1/2)^2 > (1/2) Please study these posts: ifxx2whichofthefollowingmustbetrue99506.html#p767256ifxx2whichofthefollowingmustbetrue99506.html#p767264ifxx2whichofthefollowingmustbetrue99506.html#p767437ifxx2whichofthefollowingmustbetrue99506.html#p776341ifxx2whichofthefollowingmustbetrue99506.html#p1088622ifxx2whichofthefollowingmustbetrue9950620.html#p1089273
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Re: If x<x^2 , which of the following must be true? [#permalink]
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15 Jun 2012, 00:53
Hi Bunuel, I tried to solve it this way, please let me know if this is correct as I am getting confused with the final wording of MUST be true. I am kind of stuck at the final stages to make the decision.
The problem mentions: x<x2
Case 1: if x is +ve; then x<x2, given that we do not know if x=0, then xx2<0 ==> x(1x)<0 ==> multiplying by negative sign; x(x1)>0...which means x<0 and X>1
Case 2: if x is ve; then x<x2, then xx2<0 ===>(x+x2) <0...==> x(1+x)>0 ===> x<1 and x>0
From C1 AND C2..on the number x<1 and x>1; after this I am a little lost, how to make the final call.




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