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If x < x^2, which of the following must be true?
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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 PLEASE READ THE WHOLE THREAD AND FOLLOW THE LINKS PROVIDED IN ORDER TO UNDERSTAND THE QUESTION/SOLUTIONS CORRECTLY.
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Originally posted by mehdiov on 20 Aug 2010, 12:59.
Last edited by Bunuel on 09 Apr 2018, 09:58, edited 4 times in total.
Edited the question and added the OA




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If x < x^2, which of the following must be true?
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20 Aug 2010, 13:18




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Re: If x < x^2, which of the following must be true?
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20 Aug 2010, 13:22
Hi, The XY plane of looking at it is the fastest way  I would guess. Mod(x) looks like a V x^2 is a parabola which just touches the xaxis (tangent) at the origin. When x < 1; Mod(x) < x^2 When x = 1; Mod (x) = x^2 When 1<x<0; Mod(x) > x^2 When x = 0; Mod (x) = x^2 When 0<x<1; Mod(x) > x^2 When x = 1; Mod (x) = x^2 When x > 1; Mod(x) < x^2 Hope this helps. Thanks. Wrong forum ?!?!?
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Re: If x < x^2, which of the following must be true?
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20 Aug 2010, 13:25
Bunuel wrote: III. x<1 > > may or may not be true.
Bunuel, Isn't this always true when x<x^2 ... Isn't the answer E. Thanks.
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20 Aug 2010, 13:33
4gmatmumbai wrote: Bunuel wrote: III. x<1 > > may or may not be true.
Bunuel, Isn't this always true when x<x^2 ... Isn't the answer E. Thanks. \(x<x^2\) is given as fact and then we asked to determine which of the following statements MUST be true. \(x<x^2\) means that either \(x<1\) or \(x>1\), \(x\) can be ANY value from these two ranges, (I think in your own solution you've reached this conclusion: when \(x<1\) the graph of \(x\) is below (less than) the graph of \(x^2\) and when \(x>\)1 again the graph of \(x\) is below the graph of \(x^2\)). Now, III says \(x<1\) this statement is not always true as \(x\) can be for example 3 and in this case \(x<1\) doesn't hold true. Hope it's clear.
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Re: If x < x^2, which of the following must be true?
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21 Aug 2010, 01:01
I'm confused, because if x= 2 ==> x< x^2 x< x^2, X^2 should always be > 1 Can eliminate II, because if x=1/2, x> x^2
Could you give an example Bunuel



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21 Aug 2010, 02:24
qweert wrote: I'm confused, because if x= 2 ==> x< x^2 x< x^2, X^2 should always be > 1 Can eliminate II, because if x=1/2, x> x^2
Could you give an example Bunuel Question is: "which of the following statements MUST be true", not COULD be true. We are GIVEN that \(x<x^2\), which means that either \(x<1\) OR \(x>1\). So GIVEN that: \(x<1\) OR \(x>1\). Statement II. is \(x<1\) always true? NO. As \(x\) could be more than 1, eg. 2, 3, 5.7, ... and in this case \(x<1\) is not true. So statement II which says that \(x<1\) is not always true. Hope it's clear.
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Re: If x < x^2, which of the following must be true?
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07 Sep 2010, 01:40
mehdiov, what's the OA? Bunuel, I understand your reasoning but I don't think you are right. IMO the answer is E. Should we follow your logic: Quote: Statement II. is x<1 always true? NO. As x could be more than 1, eg. 2, 3, 5.7, ... and in this case x<1 is not true. So statement II which says that x<1 is not always true. Then we can't even accept I, since x doesn't always have to be greater than 1 (it can be 10, 50 etc.). The question is similar to (but I still think that it is different): psinequality13943.htmlMaybe I don't understand something. Thanks.



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Re: If x < x^2, which of the following must be true?
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07 Sep 2010, 04:03
nonameee wrote: mehdiov, what's the OA? Bunuel, I understand your reasoning but I don't think you are right. IMO the answer is E. Should we follow your logic: Quote: Statement II. is x<1 always true? NO. As x could be more than 1, eg. 2, 3, 5.7, ... and in this case x<1 is not true. So statement II which says that x<1 is not always true. Then we can't even accept I, since x doesn't always have to be greater than 1 (it can be 10, 50 etc.).The question is similar to (but I still think that it is different): psinequality13943.htmlMaybe I don't understand something. Thanks. OA is A.\(x<x^2\) means that either \(x<1\) or \(x>1\). For example \(x\) could be 5, 3, 2, 3, ... Basically the question is: If \(x<1\) or \(x>1\) which of the following must be true? You are saying that answer is: E (I and III only). Is III: \(x<1\) ALWAYS true? Is it true for \(x=3\)? NO. So E is not correct.
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Re: If x < x^2, which of the following must be true?
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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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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: If x < x^2, which of the following must be true?
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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?
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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?
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23 May 2012, 06:09
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.
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Re: If x < x^2, which of the following must be true?
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23 May 2012, 12:09
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?
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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?
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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?
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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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Re: If x < x^2, which of the following must be true?
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15 Jun 2012, 02:59
pavanpuneet wrote: 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. First of all please study these posts: ifxx2whichofthefollowingmustbetrue99506.html#p767256ifxx2whichofthefollowingmustbetrue99506.html#p767264ifxx2whichofthefollowingmustbetrue99506.html#p767437ifxx2whichofthefollowingmustbetrue99506.html#p776341ifxx2whichofthefollowingmustbetrue99506.html#p1088622ifxx2whichofthefollowingmustbetrue9950620.html#p1089273Because the question you ask was answered before. Anyway, given that \(x<x^2\), which means that \(x<1\) or \(x>1\). Next we are given three options and are asked which of the following MUST be true. I. x^2>1. Since \(x<1\) or \(x>1\) then this option is always true (ANY number less than 1 or more than 1 when squared will be more than 1); II. x>0. This option is NOT always true, for example x could be 2 and in this case x>0 won't not true; III. x<1. This option is also NOT always true, for example x could be 2 and in this case x<1 won't not true. So, we have that only option I is always true. Answer: A (I only). Check our Must or Could be True Questions to practice more: search.php?search_id=tag&tag_id=193Hope it helps.
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