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Re: If x/|x|<x which of the following must be true about x? [#permalink]
 13 Jun 2012, 22:05
sanjoo wrote: But wat if x will be zero?? that will be infinitive..!!.. I think answer a is correct.. m still confused  . The range x > -1 does not imply that every value greater than -1 will satisfy this inequality. It implies that every value that satisfies this inequality will be greater than -1. 0 does not satisfy this inequality because the LHS is not defined for x = 0 so it is immaterial. Answer is (B).
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Bunuel wrote:
A. x<0 --> |x|=-x --> \frac{x}{-x}<x --> -1<x --> -1<x<0;
Hi Bunuel, How did you derive -1<x<0 from -1<x without further calculation (quoting from your solution quoted above)? Thanks, Diana
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dianamao wrote: Bunuel wrote:
A. x<0 --> |x|=-x --> \frac{x}{-x}<x --> -1<x --> -1<x<0;
Hi Bunuel, How did you derive -1<x<0 from -1<x without further calculation (quoting from your solution quoted above)? Thanks, Diana Hi, It is -1<x and we got this for inequality when x<0, thus combining, -1<x<0..i hope its clear.
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Re: If x/|x|<x which of the following must be true about x? [#permalink]
11 Jul 2012, 05:21
Awesome, Thank you!
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Re: If x/|x|<x which of the following must be true about x? [#permalink]
11 Jul 2012, 09:14
sanjoo wrote: VeritasPrepKarishma wrote: There was a lot of confusion between options (A) and (B). Therefore, I would like to explain why option (B) is correct using diagrams. Forget this question for a minute. Say instead you have this question: Question 1: x > 2 and x < 7. What integral values can x take? I guess most of you will come up with 3, 4, 5, 6. That’s correct. I can represent this on the number line. Attachment: Ques3.jpg You see that the overlapping area includes 3, 4, 5 and 6. Now consider this: Question 2: x > 2 or x > 5. What integral values can x take? Let’s draw that number line again. Attachment: Ques4.jpg So is the solution again the overlapping numbers i.e. all integers greater than 5? No. This question is different. x is greater than 2 OR greater than 5. This means that if x satisfies at least one of these conditions, it is included in your answer. Think of sets. AND means it should be in both the sets (i.e. overlapping). OR means it should be in at least one of the sets. Hence, which values can x take? All integral values starting from 3 onwards i.e. 3, 4, 5, 6, 7, 8, 9 … Now go back to this question. The solution is a one liner. If \frac{x}{|x|}<x which of the following must be true about x? (A) x>1(B) x>-1(C) |x|<1(D) |x|=1(E) |x|^2>1\frac{x}{|x|} is either 1 or -1. So x > 1 or x > -1 So which values can x take? All values that are included in at least one of the sets. Therefore, x > -1. But wat if x will be zero?? that will be infinitive..!!.. I think answer a is correct.. m still confused . Stating that we must have x > -1, it doesn't mean that all the numbers with this property will satisfy the inequality given in the question. Also, the question is not asking for the set of solutions of the given inequality (which means all the values for which the inequality holds). It asks for a MUST or necessary condition. And we can easily see that another necessary condition, besides B, is x being non-zero. If we assume that x\leq -1, then we get \frac{x}{|x|}=\frac{x}{-x}=-1<xwhich contradicts x\leq -1. So, necessarily x must be greater than -1. Is this enough? Are there other conditions? Do all numbers greater than -1 satisfy the given inequality in the question? This is not what the question is about. But for sure, if x is not greater than -1, than the inequality cannot hold. Therefore, x MUST be greater than -1.
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Bunuel wrote: nmohindru wrote: If \frac{x}{|x|}<x which of the following must be true about x?
(A) x>1
(B) x>-1
(C) |x|<1
(D) |x|=1
(E) |x|^2>1 This question was well explained by Durgesh and Ian Stewart, but since there are still some doubts, I'll try to add my 2 cents. First of all let's solve this inequality step by step and see what is the solution for it, or in other words let's see in which ranges this inequality holds true. Two cases for \frac{x}{|x|}<x: A. x<0 --> |x|=-x --> \frac{x}{-x}<x --> -1<x --> -1<x<0; B. x>0 --> |x|=x --> \frac{x}{x}<x --> 1<x. So given inequality holds true in the ranges: -1<x<0 and x>1. Which means that x can take values only from these ranges. ------{-1} xxxx{0}----{1} xxxxxxNow, we are asked which of the following must be true about x. Option A can not be ALWAYS true because x can be from the range -1<x<0, eg -\frac{1}{2} and x=-\frac{1}{2}<1. Only option which is ALWAYS true is B. ANY x from the ranges -1<x<0 and x>1 will definitely be more the -1, all "red", possible x-es are to the right of -1, which means that all possible x-es are more than -1. Answer: B. Bunuel just a small query here just to stick to a standard, its better not to divide by a variable as we don't know its sign, could be positive or negative so following that I did this sum , however was having difficulty at the end combining both the x ranges, please can you help \frac{x}{|x|} < xx<x|x|x-x|x|<0 x(1-|x|)<0so first case when x>0 then 1-|x|<0so 1-|x|<0 =|x|>1|x| >1 = -1 > x > 1so we have x>0and -1 > x > 1now here is my concern how to combine these two to get the final range of x I am having difficulty combining these two to get x>1 , is there any technique ? second case ( although this was easier to combine ) x(1-|x|)<0when x<0then 1-|x|>0so 1-|x|>0 = 1 >|x||x|<1 = -1< x < 1we have x<0 and -1< x < 1 now again for combining these two is there any standard way? logically i can arrive at -1<x<0
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stne wrote: Bunuel wrote: nmohindru wrote: If \frac{x}{|x|}<x which of the following must be true about x?
(A) x>1
(B) x>-1
(C) |x|<1
(D) |x|=1
(E) |x|^2>1 This question was well explained by Durgesh and Ian Stewart, but since there are still some doubts, I'll try to add my 2 cents. First of all let's solve this inequality step by step and see what is the solution for it, or in other words let's see in which ranges this inequality holds true. Two cases for \frac{x}{|x|}<x: A. x<0 --> |x|=-x --> \frac{x}{-x}<x --> -1<x --> -1<x<0; B. x>0 --> |x|=x --> \frac{x}{x}<x --> 1<x. So given inequality holds true in the ranges: -1<x<0 and x>1. Which means that x can take values only from these ranges. ------{-1} xxxx{0}----{1} xxxxxxNow, we are asked which of the following must be true about x. Option A can not be ALWAYS true because x can be from the range -1<x<0, eg -\frac{1}{2} and x=-\frac{1}{2}<1. Only option which is ALWAYS true is B. ANY x from the ranges -1<x<0 and x>1 will definitely be more the -1, all "red", possible x-es are to the right of -1, which means that all possible x-es are more than -1. Answer: B. Bunuel just a small query here just to stick to a standard, its better not to divide by a variable as we don't know its sign, could be positive or negative so following that I did this sum , however was having difficulty at the end combining both the x ranges, please can you help \frac{x}{|x|} < xx<x|x|x-x|x|<0 x(1-|x|)<0so first case when x>0 then 1-|x|<0so 1-|x|<0 =|x|>1|x| >1 = -1 > x > 1so we have x>0and -1 > x > 1now here is my concern how to combine these two to get the final range of x I am having difficulty combining these two to get x>1 , is there any technique ? second case ( although this was easier to combine ) x(1-|x|)<0when x<0 then 1-|x|>0 so 1-|x|>0 = 1 >|x||x|<1 = -1< x < 1we have x<0 and -1< x < 1 now again for combining these two is there any standard way? logically i can arrive at -1<x<0When you consider x<0, then you automatically have that |x|=-x. Similarly, when you consider x>0, then you automatically have that |x|=x. x(1-|x|)<0: x<0 --> 1-(-x)>0 --> x>-1 --> -1<x<0. x>0 --> 1-x<0 --> x>1. So, the given inequality holds for -1<x<0. and x>1. Hope it's clear.
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Re: If x/|x|<x which of the following must be true about x? [#permalink]
31 Aug 2012, 03:56
Nice!! question..It Tricks us between option A and B...I chose A in a haste when it should be B...nicely explained by Bunuel and @durgesh nice explanation for that thin line between "must be" and "always" true
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Re: If x/|x|<x which of the following must be true about x? [#permalink]
31 Aug 2012, 04:11
Nice explanation by Bunuel. It's a lot clearer now.
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nmohindru wrote: If \frac{x}{|x|}<x which of the following must be true about x?
(A) x>1
(B) x>-1
(C) |x|<1
(D) |x|=1
(E) |x|^2>1 Bunuel wrote: This question was well explained by Durgesh and Ian Stewart, but since there are still some doubts, I'll try to add my 2 cents.
First of all let's solve this inequality step by step and see what is the solution for it, or in other words let's see in which ranges this inequality holds true.
Two cases for \frac{x}{|x|}<x:
A. x<0 --> |x|=-x --> \frac{x}{-x}<x --> -1<x --> -1<x<0;
B. x>0 --> |x|=x --> \frac{x}{x}<x --> 1<x.
So given inequality holds true in the ranges: -1<x<0 and x>1. Which means that x can take values only from these ranges.
------{-1}xxxx{0}----{1}xxxxxx
Now, we are asked which of the following must be true about x. Option A can not be ALWAYS true because x can be from the range -1<x<0, eg -\frac{1}{2} and x=-\frac{1}{2}<1.
Only option which is ALWAYS true is B. ANY x from the ranges -1<x<0 and x>1 will definitely be more the -1, all "red", possible x-es are to the right of -1, which means that all possible x-es are more than -1.
Answer: B. stne wrote:
Bunuel just a small query here
just to stick to a standard, its better not to divide by a variable as we don't know its sign, could be positive or negative
so following that I did this sum , however was having difficulty at the end combining both the x ranges, please can you help
\frac{x}{|x|} < x
x<x|x|
x-x|x|<0
x(1-|x|)<0
so
first case
when x>0 then 1-|x|<0
so 1-|x|<0 =|x|>1
|x| >1 = -1 > x > 1
so we have x>0and -1 > x > 1
now here is my concern how to combine these two to get the final range of x
I am having difficulty combining these two to get x>1 , is there any technique ?
second case ( although this was easier to combine )
x(1-|x|)<0
when x<0 then 1-|x|>0
so 1-|x|>0 = 1 >|x|
|x|<1 = -1< x < 1
we have x<0 and -1< x < 1 now again for combining these two is there any standard way?
logically i can arrive at -1<x<0
Bunuel wrote:
When you consider x<0, then you automatically have that |x|=-x. Similarly, when you consider x>0, then you automatically have that |x|=x.
x(1-|x|)<0:
x<0 --> 1-(-x)>0 --> x>-1 --> -1<x<0.
[align=]x>0 --> 1-x<0 --> x>1.[/align]
So, the given inequality holds for -1<x<0. and x>1.
Hope it's clear.
Just got it Bunuel My silly query was how x>0 and x>1 translate to x>1 well I just imagined a number line and the region of overlap gave the combined equation -2 --- -1 ------0 [---------1 ---2 --3----] (i) x>0so for x>0 we have the region to the right of 0 , the area in green above for x >1 we have the area in purple below -2 ------ -1 ------0---------1 [---2 --3----] (ii) x>1so the combined equation is the region of overlap of (i) and (ii) , as shown below -2 ------ -1 ------0--------- 1 [---2 --3----] combined x>1hence x>0 and x>1 translates to x>1similarly x<0 and -1< x <1, the region of overlap gives - 1< x <0silly me ,
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Re: If x/|x|<x which of the following must be true about x? [#permalink]
01 Sep 2012, 05:33
My learning: Since this Question is asking "must be true about x" so the answer should cover the whole range of x, even if there is a gap (in this case it is from 0 to 1) If the question would have asked "what all values of x satisfy this equation", then the correct answer would be "x > 1" (considering 0 < x < 1 is missing from choices) Am I correct!?
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Re: If x/|x|<x which of the following must be true about x? [#permalink]
02 Sep 2012, 22:02
dexerash wrote: My learning:
Since this Question is asking "must be true about x" so the answer should cover the whole range of x, even if there is a gap (in this case it is from 0 to 1)
If the question would have asked "what all values of x satisfy this equation", then the correct answer would be "x > 1" (considering 0 < x < 1 is missing from choices)
Am I correct!? Yes, you are correct that 'must be true about x' means the answer should cover the entire range of x. There can be some values in that range which x cannot take. Had the question asked for the 'all values of x that satisfy this equation', the correct answer would be -1 < x< 0 or x > 1. x can take the values from -1 to 0 too so it should also appear in the range. x>1 gives only the partial range of x. Check out this post for a detailed discussion on this question: http://www.veritasprep.com/blog/2012/07 ... -and-sets/
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Re: If x/|x|<x which of the following must be true about x? [#permalink]
06 Dec 2012, 05:47
I read this technique posted here in GMATClub and it works without thinking much.
(1) Transform the equation to f(x) on one side and 0 on the other side.
x/|x| - x < 0
x(1/|x| - 1) < 0
x < 0 and |x| > 1
(2) Get check points
x < 0 ==> checkpoint: 0
|x| > 1 ==> checkpoints: 1 and -1
(3) Rearrange like this in the x-axis
+ (-1) - (0) + (1) -
<===(+)===(-1)===(-)===(0)===(+)===(1)===(-)===>
(4) Using the rule: if f(x) < 0, the values of x lie in the (-) regions
if f(x) > 0, the values of x lie in the (+) regions
(5) Answer: -1 < x < 0 and 1 < x
Thus: x is always greater than -1 ==> x> -1
Answer: B
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Re: If x/|x|<x which of the following must be true about x?
[#permalink]
06 Dec 2012, 05:47
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