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hi, A is correct... since it is given x is an integer, x can have a value of only 2 after 1.. 1 equalizes two sides , so x>2 is correct _________________

Agreed on A, but maybe I read it wrong, but without the it has to be an integer statement, A is still correct it didn't ask to solve the inequality for all values just for what is possible right?

put x = 2 in x/lxl < x => 2/l2l < 2 which holds true , but you have considered x>2

B. x>-1 => x can not be equal to -1 and also can not be equal to 0 as given => x =1 should satisfy this. but x/lxl < x does not hold true for x=1 thus not possible.

So A and B both are out.

Lets solve it.

x/lxl < x take two cases

1. \(x< 0\) => \(x/(-x) < x\)=> x>-1 for x<0 , none of the integer satisfy this. 2. \(x>0\) => \(x/x < x\)=> for x>0 x must be greater than 1 => x>1---------P

E.\(x^2 -1 >0\) =>\((x-1)*(x+1) > 0\) => either\(x > 1\) or \(x< -1\). Since x<-1 does not have any solution => x>1 is the only possibility and it matches with the equation -P

If x/|x|<x, which of the following must be true about integer x? (x is not equal to 0)

A. x > 2 B. x > -1 C. |x| < 1 D. |x| = 1 E. |x|^2 > 1

First let's solve inequality:

\(\frac{x}{|x|}< x\) 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 \(-1<x<1\) --> \(-1<x<0\) --> but as we are told that \(x\) is an integer then this range is out as there is no integer in this range.

So, finally we have that given inequality \(\frac{x}{|x|}< x\) with a restriction that \(x\) is an integer is true for: \(x>1\) --> which means that \(x\) could be: 2, 3, 4, 5, 6, ...

Now, we are asked to find out which of the following must be true about \(x\), (note the given options are not supposed to give solutions for inequality we have).

A. x > 2 --> not always true as \(x\) could be 2; B. x > -1 --> always true: we have that \(x>2\) (\(x\) is more than 2), so it must also be more than -1; C. |x| < 1 --> not true; D. |x| = 1 --> not true; E. |x|^2 > 1 --> always true: as \(x>2\) then any \(x\) from this range when squared will be more than 1.

So options B and E are always true. Question needs revision.

There is similar question at: ps-inequality-13943-40.html which does not say that \(x\) is an integer and for it the answer is B (there is my solution on page 3).

To elaborate more. Question uses the same logic as in the examples below:

If \(x=5\), then which of the following must be true about \(x\): A. x=3 B. x^2=10 C. x<4 D. |x|=1 E. x>-10

Answer is E (x>-10), because as x=5 then it's more than -10.

Or: If \(-1<x<10\), then which of the following must be true about \(x\): A. x=3 B. x^2=10 C. x<4 D. |x|=1 E. x<120

Again answer is E, because ANY \(x\) from \(-1<x<10\) will be less than 120 so it's always true about the number from this range to say that it's less than 120.

Or: If \(-1<x<0\) or \(x>1\), then 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

As \(-1<x<0\) or \(x>1\) then ANY \(x\) from these ranges would satisfy \(x>-1\). So B is always true.

\(x\) could be for example -1/2, -3/4, or 10 but no matter what \(x\) actually is it's IN ANY CASE more than -1. So we can say about \(x\) that it's more than -1.

On the other hand for example A is not always true as it says that \(x>1\), which is not always true as \(x\) could be -1/2 and -1/2 is not more than 1.

Bunnel I disagree on some points. Pleas let me know if I m wrong. Check below.

Bunuel wrote:

If x/|x|<x, which of the following must be true about integer x? (x is not equal to 0)

A. x > 2 B. x > -1 C. |x| < 1 D. |x| = 1 E. |x|^2 > 1

First let's solve inequality:

\(\frac{x}{|x|}< x\) 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 \(-1<x<1\) --> \(-1<x<0\) --> but as we are told that \(x\) is an integer then this range is out as there is no integer in this range.

So, finally we have that given inequality \(\frac{x}{|x|}< x\) with a restriction that \(x\) is an integer is true for: \(x>1\) --> which means that \(x\) could be: 2, 3, 4, 5, 6, ...

Now, we are asked to find out which of the following must be true about \(x\), (note the given options are not supposed to give solutions for inequality we have).

A. x > 2 --> not always true as \(x\) could be 2; B. x > -1 --> always true: we have that \(x>2\) (\(x\) is more than 2), so it must also be more than -1; C. |x| < 1 --> not true; D. |x| = 1 --> not true; E. |x|^2 > 1 --> always true: as \(x>2\) then any \(x\) from this range when squared will be more than 1.

So options B and E are always true. When x =2 B is not true. For all the values for which \(\frac{x}{|x|}< x\) holds true i.e. the range for x should be subset of x>2. But since x=2 does not come under x>2 B can not be always true. The question is about must be true. Thus the only option that satisfy this is E. I do not think revision is required as E should be the only correct answer. Let me know if I m wrong.

Bunnel I disagree on some points. Pleas let me know if I m wrong. Check below.

Bunuel wrote:

If x/|x|<x, which of the following must be true about integer x? (x is not equal to 0)

A. x > 2 B. x > -1 C. |x| < 1 D. |x| = 1 E. |x|^2 > 1

First let's solve inequality:

\(\frac{x}{|x|}< x\) 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 \(-1<x<1\) --> \(-1<x<0\) --> but as we are told that \(x\) is an integer then this range is out as there is no integer in this range.

So, finally we have that given inequality \(\frac{x}{|x|}< x\) with a restriction that \(x\) is an integer is true for: \(x>1\) --> which means that \(x\) could be: 2, 3, 4, 5, 6, ...

Now, we are asked to find out which of the following must be true about \(x\), (note the given options are not supposed to give solutions for inequality we have).

A. x > 2 --> not always true as \(x\) could be 2; B. x > -1 --> always true: we have that \(x>2\) (\(x\) is more than 2), so it must also be more than -1; C. |x| < 1 --> not true; D. |x| = 1 --> not true; E. |x|^2 > 1 --> always true: as \(x>2\) then any \(x\) from this range when squared will be more than 1.

So options B and E are always true. When x =2 B is not true. For all the values for which \(\frac{x}{|x|}< x\) holds true i.e. the range for x should be subset of x>2. But since x=2 does not come under x>2 B can not be always true. The question is about must be true. Thus the only option that satisfy this is E. I do not think revision is required as E should be the only correct answer. Let me know if I m wrong.

Not so, B is always true. Again, \(\frac{x}{|x|}< x\) with a restriction that \(x\) is an integer means that \(x>1\) --> \(x\) could be: 2, 3, 4, 5, 6, ...

Now, I ask you is \(x>-1\)? YES, as ANY \(x\) from the range \(x>1\) is definitely more than -1.

Is \(x>-1,000,000\)? YES. Is \(x>0.5\)? YES. Is \(x>1\)? YES. Is \(x>1.9\)? YES. All these statement are always true.

It seems that you are confused with the logic of the question, please see the examples in the end of my previous post for examples.

You are saying x >1 means x could be 2 ,3,4,5,6. Fine.

But x=2 satisfies \(\frac{x}{|x|}< x\) but not x>2.

All your examples above are correct.

What I m saying is the range of x for \(\frac{x}{|x|}< x\) should be subset for x>2. --> is this statement wrong?

You are correct when you say, if x>1 then x is definitely > -1, Because x>1 is subset of x>-1

Here the given condition is x> -1 and we are asked which of the following is always true. X>2 can not be always true because x>-1 is not the subset of x>2. But the reverse is true.

You are saying x >1 means x could be 2 ,3,4,5,6. Fine.

But x=2 satisfies \(\frac{x}{|x|}< x\) but not x>2.

All your examples above are correct.

What I m saying is the range of x for \(\frac{x}{|x|}< x\) should be subset for x>2. --> is this statement wrong?

You are correct when you say, if x>1 then x is definitely > -1, Because x>1 is subset of x>-1

Here the given condition is x> -1 and we are asked which of the following is always true. X>2 can not be always true because x>-1 is not the subset of x>2. But the reverse is true.

I hope I m not confusing you.

Why x>2 doesn't satisfy \(\frac{x}{|x|}< x\)? If \(x=3\) then \(\frac{3}{|3|}=1<3\). Also I don't get the "subset" part at all.

Anyway: Forget about \(\frac{x}{|x|}< x\) and \(x=integer\).

We have \(x=integer\) and \(x>1\) (as above means exactly that).

\(x\) could be: 2, 3, 4, 5, 6, ...It's given to be true.

Now, option B says \(x>-1\). Is it true? Is 2 more than -1? Is 3 more than -1? ... All possible \(x-es\) are more than -1. So B is always true. _________________

Also gurpreet the question asks which of the following MUST be true, or which of the following is ALWAYS true. For such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

So can you give me an example of \(x\) which satisfies \(\frac{x}{|x|}< x\) and is not more than -1? _________________

x>-1 and x not equal to 0 and 1 - This is given. => x could be 2 , 3, 4,5,6,7----- so on Every value of x should satisfy the answer.

Q1. Which of the following is always true.

1. X>2 -> is this always true? -> This is always true expect for x = 2.

If I take x = 2 using the inequality x/|x| < x , x =2 does not satisfy x>2.

I hope it makes sense.

Bunnel my point is every value that satisfy x/|x| < x should satisfy the answer choice.

Since 2 satisfies x/|x| < x but no x>2, x>2 can not be the answer...

Subset reason : All the values of x that satisfy x/|x| < x should be a subset of x>2 => if x/|x| < x satisfy n values, then all those n values should be satisfied by x>2. _________________

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