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Re: If x<x^2 , which of the following must be true? [#permalink]
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23 Apr 2014, 02:09
PathFinder007 wrote: Bunuel wrote: 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. Hi bunnel, Can we write this as x<0 or x>0 instead of x<1 or x>1 Please clarify. No, \(x<x^2\) does not hold true if x is a negative or positive fraction. For example, if x=1/2, then (x=1/2) > (x^2=1/4).
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Re: If x<x^2 , which of the following must be true? [#permalink]
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09 May 2014, 23:31
If x≠0 and xx<x, which of the following must be true? A. x>1 B. x>−1 C. x<1 D. x>1 E. −1<x<0
we do get a range of x i.e. 1<x<0 & x>1. However, Option B cannot be always true
For instance, if 0<x<1, in this range  the inequality will not hold true. i.e. (1/2)/1/2 is not less than 1/2 (substituted the value of x  1/2 in the inequality)
Please have a look at this and let me know in case I am missing out something.



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Re: If x<x^2 , which of the following must be true? [#permalink]
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10 May 2014, 04:21



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Re: If x<x^2 , which of the following must be true? [#permalink]
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06 Jul 2014, 22:24
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 Bunnel, I have selected E as the ans. I used the number to solve this question. Definately x <x^2 if x <1 or x>1. Just want to know why you are telling x<1 is not true?. If we take anv value that is l.t. 1 then x < x^2. Then why we are not choosing E? Thanks.



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Re: If x<x^2 , which of the following must be true? [#permalink]
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07 Jul 2014, 00:31
PathFinder007 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). Hi Bunnel, I have selected E as the ans. I used the number to solve this question. Definately x <x^2 if x <1 or x>1. Just want to know why you are telling x<1 is not true?. If we take anv value that is l.t. 1 then x < x^2. Then why we are not choosing E? Thanks. There are 4 pages of discussion of this question. A lot of useful staff on those pages including answer to your question. Please read.
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Re: If x<x^2 , which of the following must be true? [#permalink]
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07 Jul 2014, 10:53
Hi Bunuel, Can you please direct me to the discussion page? you mentioned: There are 4 pages of discussion of this question. A lot of useful staff on those pages including answer to your question. Please read.[/quote]
am not able to go to that discussion from this page. Bunuel wrote: GmatDestroyer2013 wrote: If x≠0 and xx<x, which of the following must be true? A. x>1 B. x>−1 C. x<1 D. x>1 E. −1<x<0
we do get a range of x i.e. 1<x<0 & x>1. However, Option B cannot be always true
For instance, if 0<x<1, in this range  the inequality will not hold true. i.e. (1/2)/1/2 is not less than 1/2 (substituted the value of x  1/2 in the inequality)
Please have a look at this and let me know in case I am missing out something. The ranges for which xx<x is true are x<1 and x>1. Check the discussion on previous pages.
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Re: If x<x^2 , which of the following must be true? [#permalink]
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07 Jul 2014, 11:17



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Re: If x<x^2 , which of the following must be true? [#permalink]
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08 Jul 2014, 08:00
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?
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Re: If x<x^2 , which of the following must be true? [#permalink]
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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? [#permalink]
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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? [#permalink]
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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? [#permalink]
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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? [#permalink]
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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? [#permalink]
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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? [#permalink]
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15 Jul 2015, 08:46
Hello Bunuel, With regards to statement 3 of this ques ( x<1). i drew the graphs of both the functions ( x and x^2). For x<1 , in the graph , i am able to see that x^2 is always greater than x . Then how is statement 3 correct ? I have read all the posts above, pertaining to this particular problem but unable to understand this. Please help. Thanks in advance.



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Re: If x<x^2 , which of the following must be true? [#permalink]
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24 Sep 2015, 01:14
hi Bunuel,
i think most of the people are getting confused in this problem, even i picked option E.
here after solving the inequality, we get the range i.e. solution for this question. and now from the options given, we need to select that option which includes the whole range i.e. x< 1 and x>1. once we understand this, question becomes easy.
correct me if i am wrong.



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Re: If x<x^2 , which of the following must be true? [#permalink]
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24 Sep 2015, 01:15
hi Bunuel,
i think most of the people are getting confused in this problem, even i picked option E.
here after solving the inequality, we get the range i.e. solution for this question. and now from the options given, we need to select that option which includes the whole range i.e. x< 1 and x>1. once we understand this, question becomes easy.
correct me if i am wrong.



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Re: If x<x^2 , which of the following must be true? [#permalink]
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03 Oct 2015, 03:55
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, Doubt regarding third statement. The expression in the question is valid for all x>1 and X<1 As per this even third option should be right. Also, please tell me how to identify these options as in some questions, a part range is also among the answers while in others it is not. I always get confused in these questions. Is it because of "must" or "could"? Could you please mention similar questions where I can practice for this?



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Re: If x<x^2 , which of the following must be true? [#permalink]
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03 Oct 2015, 06:08
longfellow 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 ? Hi, Doubt regarding third statement. The expression in the question is valid for all x>1 and X<1 As per this even third option should be right. Also, please tell me how to identify these options as in some questions, a part range is also among the answers while in others it is not. I always get confused in these questions. Is it because of "must" or "could"? Could you please mention similar questions where I can practice for this? "Must" Means It will be true for all possible values of x "Could" Means It may be true for some values of x and may not be true for other values of x About the third: III. x<1 Take some values of x less than 1 and check if the Inequation is satisfied @x = 2, x<x^2 this is satisfied @x = 3, x<x^2 this is satisfied as well for all values of x less than 1 xwill have positive and smaller absolute value than absolute values of x^2 which will be positive. x less than 1 is an acceptable range of values of x but since x may be positive (greater than 1) as well so it is not necessary that x be less than 1 I hope this helps!
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Re: If x<x^2 , which of the following must be true? [#permalink]
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03 Oct 2015, 22:58
GMATinsight wrote: longfellow 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 ? Hi, Doubt regarding third statement. The expression in the question is valid for all x>1 and X<1 As per this even third option should be right. Also, please tell me how to identify these options as in some questions, a part range is also among the answers while in others it is not. I always get confused in these questions. Is it because of "must" or "could"? Could you please mention similar questions where I can practice for this? "Must" Means It will be true for all possible values of x "Could" Means It may be true for some values of x and may not be true for other values of x About the third: III. x<1 Take some values of x less than 1 and check if the Inequation is satisfied @x = 2, x<x^2 this is satisfied @x = 3, x<x^2 this is satisfied as well for all values of x less than 1 xwill have positive and smaller absolute value than absolute values of x^2 which will be positive. x less than 1 is an acceptable range of values of x but since x may be positive (greater than 1) as well so it is not necessary that x be less than 1 I hope this helps! Thanks GMATInsight. I would still like practice on more questions like this one? Could you point out a few?




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