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Re: If x < x^2, which of the following must be true?
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16 Feb 2012, 22:02
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: If x < x^2, which of the following must be true?
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16 Feb 2012, 22:06
devinawilliam83 wrote: 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?
x^21>0 > x<1 or x>1. Solving inequalities: x24x94661.html#p731476inequalitiestrick91482.htmldatasuffinequalities109078.htmlrangeforvariablexinagiveninequality109468.html?hilit=extreme#p873535everythingislessthanzero108884.html?hilit=extreme#p868863Hope it helps.
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Re: If x < x^2, which of the following must be true?
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21 May 2012, 07:28
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, 08:00
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, 04:10
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, 11:09
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, 20:16
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, 01:01
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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14 Jun 2012, 23:53
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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20 Jun 2012, 05:41
Hi, I am having issues with a fundanmental concept. In the below explanation x(x+1)>0 , isnt x>0 and x>1 (from x+1>0) rather than x<1 and x>0. clearly by inspection my answer is wrong, but was confused over how you go correct answer (is it possible to show the steps)
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; appreciate your input Thanks



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Re: If x < x^2, which of the following must be true?
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21 Jun 2012, 18:15
Hi Krishma, the link that you posted is probably best for summary. It ends with the comment "Anyone wants to add ABSOLUTE VALUES....That will be a value add to this post ". Can you suggest any such thread where absolute value side is summarized like this as well. Appreciate your input Thanks



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Re: If x < x^2, which of the following must be true?
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28 Oct 2012, 02:31
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). Hi bunuel, sorry , but I didn't undestrand how you reduce x<x^2 by x to get the solution...x<1 and x>1 please, could you explain me in more details? Thanks



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Re: If x < x^2, which of the following must be true?
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09 Mar 2013, 23:42
Got the answer from : ifxx2whichofthefollowingmustbetrue9950620.html#p1098244But can some one draw the graph
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Re: If x < x^2, which of the following must be true?
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11 Mar 2013, 00:21
greatps24 wrote: x will be less than x^2 in the region where the graph of x is below the graph of x^2. Graph of y = x is a V at (0, 0) and graph of y = x^2 is an upward facing parabola at (0, 0) Attachment:
Ques3.jpg [ 7.72 KiB  Viewed 2932 times ]
Just looking at the graph should tell you that answer must be (A). You don't need to solve. x will lie in one of two ranges x > a or x < a. II and III will not work in any case. I must work since one of them has to be true according to the options.
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Re: If x < x^2, which of the following must be true?
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08 Jul 2014, 08:24
bankerboy30 wrote: Here's the way I look at it:
I think were all in agreement that the conditions for the statement is x<1 and x>1. From this we can conclude this statement is true if we have values of x<1 or x>1. But for all values to be true if I plug a value of x<1 or x>1 into the answer choice inequalities then the inequality must hold up.
So looking at my option choices:
First option: x^2 > 1. Let's pick a number less than 1. Say 2. Is (2)^2 > 1? Yes. Now let's look for a number x>1. Say 2. Is (2)^2>1. Yes. So this holds true for all of our conditions of x (x>1, x<1).
Second option: x>0. Let's pick the same numbers we picked before. 2,2. Is 2>0. Yes. Is 2>0. No this won't work if x<1 or x is between 0 and 1.
Third option: x<1. Again let's use the same numbers that we have been using to validate all possible solutions. Say x is 2. Is 2<1? No. Is 2<1? Yes. However both don't satisfy the equation
The reason why the first one does is because of the x^2. When were squaring both a positive and negative number that is x<1 or x>1 then the result will always be a positive number that is greater than one.
Bunuel does this interpretation make sense?
Posted from my mobile device Yes, that's correct. From the stem we know that x is some number either less than 1 or more than 1. Which of the statement is true about that number? I says that the square of that number is greater than 1. Well, square of ANY number less than 1 or more than 1 is greater than 1. So, no matter what x actually is, this statement is always true. II says that that number is greater than 0. That might not be true if it's less than 1. III says that that number is greater less than 1. That might not be true if it's more than 1. So, only statement I is 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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13 Jun 2015, 11:07
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. Hi Bunuel, I have a small doubt. Can we solve the obve equation like this: if x>0, \(x^2 > x\) (now dividing both the sides by x) we get x > 1 (this is one range) the other range is when x <0 and x =x, we get \(x^2 > x\) (dividing by x) x>1 so the range is x>1 and x>1. where am i going wrong?



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Re: If x < x^2, which of the following must be true?
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13 Jun 2015, 11:49
arshu27 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. Hi Bunuel, I have a small doubt. Can we solve the obve equation like this: if x>0, \(x^2 > x\) (now dividing both the sides by x) we get x > 1 (this is one range) the other range is when x <0 and x =x, we get \(x^2 > x\) (dividing by x)
x>1
so the range is x>1 and x>1.where am i going wrong? x>1 and x>1 does not make any sense. As for the mistake you are making, when dividing x^2 > x by x, which is negative, you should flip the sign: x < 1.
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Re: If x < x^2, which of the following must be true?
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27 Jun 2015, 10:36
Hi, i wonder can we solve this inequality like this square each side thus we would have x^2<x^4 or X^2(x1)(x+1)<0, thus we have following solution x (1:0) and (0:1). So based on that NONE is the answer. Please correct me or am i making some fundamental error.



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Re: If x < x^2, which of the following must be true?
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28 Jun 2015, 22:01
mika84 wrote: Hi, i wonder can we solve this inequality like this square each side thus we would have x^2<x^4 or X^2(x1)(x+1)<0, thus we have following solution x (1:0) and (0:1). So based on that NONE is the answer. Please correct me or am i making some fundamental error. x^2 < x^4 x^2  x^4 < 0 x^2 (1  x^2) < 0 x^2(x^2  1) > 0 (multiplying by 1 and hence flipping the inequality sign) x^2(x  1)(x + 1) > 0 The solution to this is x < 1 or x > 1. You ignore x^2 because it has even power. The transition points are only 1 and 1. Check this post for more: http://www.veritasprep.com/blog/2012/07 ... spartii/
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Re: If x < x^2, which of the following must be true?
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28 Jun 2015, 23:11
mika84 wrote: Hi, i wonder can we solve this inequality like this square each side thus we would have x^2<x^4 or X^2(x1)(x+1)<0, thus we have following solution x (1:0) and (0:1). So based on that NONE is the answer. Please correct me or am i making some fundamental error. x^2 < x^4 Since the power is even both sides so both sides must be positive which has two inferences 1) The range of Negative values of x will be valid along with the same range of positive values of x 2) Higher powers of x are greater than lower powers of x only when x is greater than 1 (for positive values of x) I.e. the Required range must be x >1 or x < 1ALTERNATIVELYWe can solve the inequation, x^2  x^4 < 0 x^2 (1  x^2) < 0 x^2(x^2  1) > 0 (multiplying by 1 and hence flipping the inequality sign) x^2(x  1)(x + 1) > 0 Since x^2 is positive for all values of x except zero therefore (x  1)(x + 1) > 0 I.e. both the parts must be either positive or both must be Negative For both of (x1) and (x+1) to be positive, x must be greater than 1 For both of (x1) and (x+1) to be negative, x must be less than 1 The solution to this is x < 1 or x > 1.
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Re: If x < x^2, which of the following must be true?
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