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KarishmaB
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If x/|x|<x which of the following must be true about x? 12 Jun 2012, 04:31
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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
Ques3.jpg [ 4.45 KiB | Viewed 34107 times ]
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
Ques4.jpg [ 4.24 KiB | Viewed 34109 times ]
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.
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If x/|x|<x which of the following must be true about x? 13 Jun 2012, 13:04
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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.
Show more


But wat if x will be zero?? that will be infinitive..!!..
I think answer a is correct.. m still confused :(.
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If x/|x|<x which of the following must be true about x? 13 Jun 2012, 22:05
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sanjoo

But wat if x will be zero?? that will be infinitive..!!..
I think answer a is correct.. m still confused :(.
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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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If x/|x|<x which of the following must be true about x? 11 Jul 2012, 09:14
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VeritasPrepKarishma
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 :(.
Show more

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 \(-1\).
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If x/|x|<x which of the following must be true about x? 01 Sep 2012, 05:33
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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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If x/|x|<x which of the following must be true about x? Updated on: 11 Oct 2022, 02:13
Originally posted by KarishmaB on 02 Sep 2012, 22:02.
Last edited by KarishmaB on 11 Oct 2022, 02:13, edited 1 time in total.
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dexerash
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!?
Show more

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.
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If x/|x|<x which of the following must be true about x? 24 Mar 2014, 07:40
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Hi Bunuel,
Can you help clear the confusion with X = 1/2.

Thanks.
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If x/|x|<x which of the following must be true about x? 24 Mar 2014, 07:51
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seabhi
Hi Bunuel,
Can you help clear the confusion with X = 1/2.

Thanks.
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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

We are given that \(\frac{x}{|x|}<x\) (this is a true inequality), so first of all we should find the ranges of \(x\) for which this inequality holds true.

\(\frac{x}{|x|}< x\) multiply both sides of inequality by \(|x|\) (side note: we can safely do this as absolute value is non-negative and in this case we know it's not zero too) --> \(x<x|x|\) --> \(x(|x|-1)>0\):
Either \(x>0\) and \(|x|-1>0\), so \(x>1\) or \(x<-1\) --> \(x>1\);
Or \(x<0\) and \(|x|-1<0\), so \(-1<x<1\) --> \(-1<x<0\).

So we have that: \(-1<x<0\) or \(x>1\). Note \(x\) is ONLY from these ranges.

Option B says: \(x>-1\) --> ANY \(x\) from above two ranges would be more than -1, so B is always true.

Answer: B.
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If x/|x|<x which of the following must be true about x? 07 Jul 2014, 01:32
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nmohindru
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.
Show more

Bunnel, if I were to multiply the original stem with |x| (since |x| is always positive) it would result in x*(|x|-1) > 0.
This would mean x > 0 and |x| > 1
|x| > 1 would lead to x < -1 and x > 1 . This is completely different from the answer you've reached. I see that your method is accurate and the answer justified, but can you please correct my method here.
Thanks in advance!
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If x/|x|<x which of the following must be true about x? 07 Jul 2014, 01:41
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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.

Bunnel, if I were to multiply the original stem with |x| (since |x| is always positive) it would result in x*(|x|-1) > 0.
This would mean x > 0 and |x| > 1
|x| > 1 would lead to x < -1 and x > 1 . This is completely different from the answer you've reached. I see that your method is accurate and the answer justified, but can you please correct my method here.
Thanks in advance!
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It would give the same answer.

\(x*(|x|-1) > 0\). This implies that both multiples have the same sign.

\(x>0\) and \(|x|>1\) (since we consider positive x, then this transforms to x>1) --> \(x>1\).
\(x<0\) and \(|x|<1\) (since we consider negative x, then this transforms to -x<1 --> -1<x) --> \(-1<x<0\).

The same ranges as in my solution.

Hope it's clear.
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If x/|x|<x which of the following must be true about x? 15 Aug 2014, 07:50
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Bunuel,

What if we square the inequality x/ |x| < x. Then we get (x^2 / x ) < x^2 which implies that x^2 < x^3. Is this correct? Please explain. Thanku
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If x/|x|<x which of the following must be true about x? 15 Aug 2014, 09:28
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sri30kanth
Bunuel,

What if we square the inequality x/ |x| < x. Then we get (x^2 / x ) < x^2 which implies that x^2 < x^3. Is this correct? Please explain. Thanku
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We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality), which is not the case here.

Also, the second step in your solution: never multiply (or reduce) an inequality by a variable (or the expression with a variable) if you don't know the sign of it, we don't know the sign of x, so we cannot multiply x^2/x < x^2 by x here.

For more check here: inequalities-tips-and-hints-175001.html
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If x/|x|<x which of the following must be true about x? 22 Apr 2015, 19:22
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Brunel, Can you please explain why option E is not feasible?
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If x/|x|<x which of the following must be true about x? 23 Apr 2015, 01:44
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AverageGuy123
Brunel, Can you please explain why option E is not feasible?
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Dear AverageuGuy123

As Bunuel explained above,

Either -1 < x < 0
Or x > 1

Now, |x| as you know, represents the magnitude of x. Option E says that |x|^2 must be greater than 1.

Let's first consider the case when -1 < x < 0

A possible value of x in this case is -0.5
So, what is the value of |x|^2? It is equal to 0.25

Is it greater than 1? NO

Let's now consider the case when x > 1

A possible value of x in this case is 2.
So, what is the value of |x|^2? It's 4.

Is it greater than 1? YES

So, as we see, that |x|^2 CAN BE greater than 1. But can we say that |x|^2 MUST BE greater than 1? NO, because |x|^2 is not greater than 1 for all possible values of x.

So, the key takeaway from this discussion is that:

we need to be careful whether the question is asking about MUST BE TRUE statements or about CAN BE TRUE statements.

Hope this helped! :)

- Japinder
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If x/|x|<x which of the following must be true about x? 23 Apr 2015, 04:36
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Brunel, Can you please explain why option E is not feasible?
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You can plug in numbers to eliminate options.

"which of the following must be true about x" means that every acceptable value of x must lie in the range given in the correct option. The acceptable values of x are the values for which x/|x| < x.

A. x>2
Must x be greater than 2?

This should make you check for 2.
2/|2| < 2
1 < 2 (True)
So 2 is an acceptable value of x. But 2 is not greater than 2.
So this option is not correct. This also makes you eliminate options (C) and (D).

E. |x|^2>1
Must x be greater than 1 or less than -1?

Check for 1/2
(1/2)/|1/2| < 1/2
1 < 1/2 (False)

Check for -1/2
(-1/2)/|-1/2| < -1/2
-1 < -1/2 (True)

So x = -1/2 is an acceptable value but it does not lie in this range. Hence option (E) is also incorrect.

Answer must be (B)
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If x/|x|<x which of the following must be true about x? 26 Sep 2020, 12:55
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shanky315
What is wrong with Option A, plugging numbers satisfy this condition. Bunuel can you please suggest.

Posted from my mobile device
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x can be -1/2 and in this case x > 1 (option A) will not be correct.
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If x/|x|<x which of the following must be true about x? 16 Nov 2023, 05:34
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Bunuel
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

We are given that \(\frac{x}{|x|}<x\) (this is a true inequality), so first of all we should find the ranges of \(x\) for which this inequality holds true.

\(\frac{x}{|x|}< x\) multiply both sides of inequality by \(|x|\) (side note: we can safely do this as absolute value is non-negative and in this case we know it's not zero too) --> \(x<x|x|\) --> \(x(|x|-1)>0\):
Either \(x>0\) and \(|x|-1>0\), so \(x>1\) or \(x<-1\) --> \(x>1\);
Or \(x<0\) and \(|x|-1<0\), so \(-1<x<1\) --> \(-1<x<0\).

So we have that: \(-1<x<0\) or \(x>1\). Note \(x\) is ONLY from these ranges.

Option B says: \(x>-1\) --> ANY \(x\) from above two ranges would be more than -1, so B is always true.

Answer: B.
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When we take the negative value for x won't it be -x/|x| <-x , why are we not considering the negative sign on the right side ? So once we solve it we get 1>x. Am I missing something?
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If x/|x|<x which of the following must be true about x? 16 Nov 2023, 06:47
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benetamary
Bunuel
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

We are given that \(\frac{x}{|x|}<x\) (this is a true inequality), so first of all we should find the ranges of \(x\) for which this inequality holds true.

\(\frac{x}{|x|}< x\) multiply both sides of inequality by \(|x|\) (side note: we can safely do this as absolute value is non-negative and in this case we know it's not zero too) --> \(x<x|x|\) --> \(x(|x|-1)>0\):
Either \(x>0\) and \(|x|-1>0\), so \(x>1\) or \(x<-1\) --> \(x>1\);
Or \(x<0\) and \(|x|-1<0\), so \(-1<x<1\) --> \(-1<x<0\).

So we have that: \(-1<x<0\) or \(x>1\). Note \(x\) is ONLY from these ranges.

Option B says: \(x>-1\) --> ANY \(x\) from above two ranges would be more than -1, so B is always true.

Answer: B.
When we take the negative value for x won't it be -x/|x| <-x , why are we not considering the negative sign on the right side ? So once we solve it we get 1>x. Am I missing something?
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If x is negative, it's incorrect to replace x with -x. For instance, consider x = -2. In this case, x is already negative. There's no logical reason to alter it to -x = -2, as it would distort the original value.
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If x/|x|<x which of the following must be true about x? 28 Nov 2023, 00:02
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Bunuel
nmohindru
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 cannot 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.
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Hi Bunuel ,

While we choose x>-1, this will hold true for x=1 as well.
Then, 1/1 is not < 1
1/1 = 1

The equation will be satisfied for all values of x>1.

Therefore, I think A shall be the answer. I do understand that in situations such as x=-1/2, the equation is satisfied. However, option B does not seem to be full proof.

Let me know what you think.
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If x/|x|<x which of the following must be true about x? 28 Nov 2023, 00:09
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Gprabhumir
Bunuel
nmohindru
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 cannot 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.

Hi Bunuel ,

While we choose x>-1, this will hold true for x=1 as well.
Then, 1/1 is not < 1
1/1 = 1

This logic will hold true for all values of x>1.

Therefore, I think A shall be the answer. I do understand that in situations such as x=-1/2, the equation is satisfied. However, option B does not seem to be full proof.

Let me know what you think.
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I believe your question has been addressed several times in this thread. To avoid repetition, I recommend rereading the solution carefully. I hope it helps.
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