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Re: If x/|x|<x which of the following must be true about x? [#permalink]
13 Jun 2012, 21:05

Expert's post

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). _________________

Re: If x/|x|<x which of the following must be true about x? [#permalink]
11 Jul 2012, 08:14

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 \(\frac{x}{|x|}=\frac{x}{-x}=-1<x\) which 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 \(-1\). _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

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}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.

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 \(x>0\)and \(-1 > x > 1\)

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 \(-1<x<0\) _________________

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}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.

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 \(x>0\)and \(-1 > x > 1\)

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 \(-1<x<0\)

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.

Re: If x/|x|<x which of the following must be true about x? [#permalink]
31 Aug 2012, 02:56

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

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 \(x>0\)and \(-1 > x > 1\)

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.

Re: If x/|x|<x which of the following must be true about x? [#permalink]
01 Sep 2012, 04:33

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)

Re: If x/|x|<x which of the following must be true about x? [#permalink]
02 Sep 2012, 21:02

Expert's post

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.

Re: If x/|x|<x which of the following must be true about x? [#permalink]
30 May 2013, 21:32

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.

Hi Karishma/Bunnel,

Had the question been 'Which of the values of x satisfies this inequality?' then the correct answer would have been A

Now the question is essentially, 'Which of the following ranges consists of all values of x that satisfy this inequality?'

Re: If x/|x|<x which of the following must be true about x? [#permalink]
31 May 2013, 01:52

Expert's post

cumulonimbus 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.

Hi Karishma/Bunnel,

Had the question been 'Which of the values of x satisfies this inequality?' then the correct answer would have been A

Now the question is essentially, 'Which of the following ranges consists of all values of x that satisfy this inequality?'

Re: If x/|x|<x which of the following must be true about x? [#permalink]
09 Jul 2013, 21:34

If x/|x|<x which of the following must be true about x?

This can be solved by plugging in numbers from each of the five answer choices:

(A) x>1 x/|x|<x 2/|2| < 2 1<2 Valid - this is a possible answer choice

(B) x>-1 x/|x|<x -0.5/|-0.5|<-0.5 -1<-0.5 Valid - this is a possible answer choice

(C) |x|<1 -1<x<1 x/|x|<x 0/|0|<0 0<0 Invalid - 0 is not less than 0

(D) |x|=1 x=1, x=-1 x/|x|<x 1/|1|<1 1<1 Invalid as 1 is not less than 1

(E) |x|^2>1 x>1, x<-1 x/|x|<x -2/|-2|<-2 -1< -2 Invalid as -1 is GREATER than -2

While (a) is correct, (b) is the right answer as it includes valid possibilities that are not included in (a)

(B)

Also, solving algebraically: x/|x|<x Two cases: x>0 x/|x|<x x/x<x 1<x (If x>0 and x>1 then the intersection is at x>1 because only x>1 contains all cases that x>0 does) x<0 x/|x|<x x/(-x)<x -1<x -1<x<0

So: -1<x<0 AND 1<x Use that with the answer choices to find the correct answer.

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...