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Re: If x < x^2, which of the following must be true? [#permalink]
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20 Jun 2012, 06:41
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? [#permalink]
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20 Jun 2012, 06:44



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Re: If x < x^2, which of the following must be true? [#permalink]
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20 Jun 2012, 22:12
idreesma wrote: 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 Responding to a pm: The problem you are facing is that you do not know how to handle inequalities. How do you get the range for which this inequality holds? x(x+1) > 0 Think of it this way: Product of x and (x+1) should be positive. When will that happen? When either both the terms are positive or both are negative. Case 1: When both are positive x > 0 x + 1 > 0 i.e. x > 1 For both to be positive, x must be greater than 0. Hence this inequality will hold when x > 0. Case 2: When both are negative x < 0 x + 1 < 0 i.e. x < 1 For both to be negative, x must be less than 1. Hence this inequality will hold when x < 1. So we get two ranges in which this inequality holds: x > 0 or x < 1. The fastest way to solve it is using the number line. Check this post for the explanation of this method: inequalitiestrick91482.htmlAlso, your Veritas book discusses this concept too.
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Re: If x < x^2, which of the following must be true? [#permalink]
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21 Jun 2012, 19:15
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? [#permalink]
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04 Oct 2012, 21:21
idreesma wrote: 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 I have discussed mods here: http://www.veritasprep.com/blog/2011/01 ... edoredid/http://www.veritasprep.com/blog/2011/01 ... htomods/http://www.veritasprep.com/blog/2011/01 ... spartii/
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Re: If x < x^2, which of the following must be true? [#permalink]
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28 Oct 2012, 03:31
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? [#permalink]
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29 Oct 2012, 02:51
mario1987 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 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 Express \(x^2\) as \(x*x\), so we have that: \(x<x*x\) > reduce by \(x\) (notice that \(x\) is positive, thus we can safely divide both parts of the inequality by it) > \(1<x\) > \(x<1\) or \(x>1\). 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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10 Mar 2013, 00:42
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? [#permalink]
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10 Mar 2013, 02:01
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X < x^2 means X is not a decimal. means X < 1 or x > 1. Square of anything will be positive. Hence. A Hope it helps.



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Re: If x < x^2, which of the following must be true? [#permalink]
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11 Mar 2013, 01:21
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 1666 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? [#permalink]
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21 Jul 2013, 20:25
1. x<x^2 => (i) x < x^2 or (ii) x < x^2 2. (1) => x>1 or x<1 . Only this satisfies both (i) and (ii) Is x^2>1 always true? Yes follows from (2) Is x>0 always true, No because x could be <1 Is x<1 always true?, No because x could be >1 So answer is choice A.
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Re: If x < x^2, which of the following must be true? [#permalink]
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08 Jul 2014, 09:00
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?
Posted from my mobile device



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Re: If x < x^2, which of the following must be true? [#permalink]
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08 Jul 2014, 09:24
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, 12:07
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, 12:49
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, 11:36
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, 23:01
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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29 Jun 2015, 00:11
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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03 Oct 2015, 04:55
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, 07:08
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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