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Guys, I do understand how to do it through plugging in numbers, can someone elaborate the method, where you use algebraic method. It would be great help! Thank you, I am considering two options x>0 and x<0 1) x>0 then cross multiplying we get x<x^2 then we find out that x is >1

2) x<0 then cross multiplying we get that x>-x^2 then we get that x is between -1 and 0 thus, x is between -1 and 0 and x>1

Guys, I do understand how to do it through plugging in numbers, can someone elaborate the method, where you use algebraic method. It would be great help! Thank you, I am considering two options x>0 and x<0 1) x>0 then cross multiplying we get x<x^2 then we find out that x is >1

2) x<0 then cross multiplying we get that x>-x^2 then we get that x is between -1 and 0 thus, x is between -1 and 0 and x>1

If \(\frac{x}{|x|} \lt x\) , which of the following must be true about \(x\)? (\(x \ne 0\)) A. \(x\gt 2\) B. \(x \in (-1,0) \cup (1,\infty)\) C. \(|x| \lt 1\) D. \(|x| = 1\) E. \(|x|^2 \gt 1\)

Absolute value properties: Absolute value is always non-negative: \(|x|\geq{0}\) (not positive but non-negative, meaning that absolute value can equal to zero), so: When \(x\leq{0}\) then \(|x|=-x\) (note that in this case \(|x|=-negative=positive\)); When \(x\geq{0}\) then \(|x|=x\).

B. Substitute numbers. I chose 3, -3, -1/3 & 1/3 for 3 & -1/3 the inequality holds good. Instead of 3 we can have any number between 1 & infinity. Similarly instead of -1/3 we can have any number between 0 & -1

Hi all, i disagree with most of you. My Answer is A. Explanation:- x/|x| <x Case 1: if x<0 If x<0, say a no -3. Then, -3/|-3|<-3 -3/3<-3 -1<-3 which is not true. Case 2: x=0 Not applicable, as per the given condition. Case 3: if x>0 Take a no say 1(x>0) Then, 1/|1|<1 1/1<1 1<1 which is not true. Take another no say 2(x>0) Then, 2/|2|<2 2/2<2 1<2 which is true. So , for a Question like which of the following must be true about ? ( )

The answer is X>2, which is Option A.
_________________

kudos me if you like my post.

Attitude determine everything. all the best and God bless you.

Hi Dauren, Here is my solution. My ANS is A. Explanation:- x/|x| <x Case 1: if x<0 If x<0, say a no -3. Then, -3/|-3|<-3 -3/3<-3 -1<-3 which is not true. Case 2: x=0 Not applicable, as per the given condition. Case 3: if x>0 Take a no say 1(x>0) Then, 1/|1|<1 1/1<1 1<1 which is not true. Take another no say 2(x>0) Then, 2/|2|<2 2/2<2 1<2 which is true. So , for a Question like which of the following must be true about ?

The answer is X>2, which is Option A.

!!!Give me Kudos if you like my post.!!!!
_________________

kudos me if you like my post.

Attitude determine everything. all the best and God bless you.

Hi all, i disagree with most of you. My Answer is A. Explanation:- x/|x| <x Case 1: if x<0 If x<0, say a no -3. Then, -3/|-3|<-3 -3/3<-3 -1<-3 which is not true. Case 2: x=0 Not applicable, as per the given condition. Case 3: if x>0 Take a no say 1(x>0) Then, 1/|1|<1 1/1<1 1<1 which is not true. Take another no say 2(x>0) Then, 2/|2|<2 2/2<2 1<2 which is true. So , for a Question like which of the following must be true about ? ( )

The answer is X>2, which is Option A.

Number plugging is not the best method to solve this question.

OA for this question is B, not A: the given inequality holds true in two ranges \(-1<x<0\) and \(x>1\) (see solution in my previous post). You can try values from this ranges to check. So x>2 is not always true.
_________________

Hi all, i disagree with most of you. My Answer is A. Explanation:- x/|x| <x Case 1: if x<0 If x<0, say a no -3. Then, -3/|-3|<-3 -3/3<-3 -1<-3 which is not true. Case 2: x=0 Not applicable, as per the given condition. Case 3: if x>0 Take a no say 1(x>0) Then, 1/|1|<1 1/1<1 1<1 which is not true. Take another no say 2(x>0) Then, 2/|2|<2 2/2<2 1<2 which is true. So , for a Question like which of the following must be true about ? ( )

The answer is X>2, which is Option A.

Number plugging is not the best method to solve this question.

OA for this question is B, not A: the given inequality holds true in two ranges \(-1<x<0\) and \(x>1\) (see solution in my previous post). You can try values from this ranges to check. So x>2 is not always true.

x>2 will alwayz be true as it is a part of (-1,0)& (1,infinity) anyway, i realized my mistake. the answer will be B. thnx buddy!
_________________

kudos me if you like my post.

Attitude determine everything. all the best and God bless you.

Hi all, i disagree with most of you. My Answer is A. Explanation:- x/|x| <x Case 1: if x<0 If x<0, say a no -3. Then, -3/|-3|<-3 -3/3<-3 -1<-3 which is not true. Case 2: x=0 Not applicable, as per the given condition. Case 3: if x>0 Take a no say 1(x>0) Then, 1/|1|<1 1/1<1 1<1 which is not true. Take another no say 2(x>0) Then, 2/|2|<2 2/2<2 1<2 which is true. So , for a Question like which of the following must be true about ? ( )

The answer is X>2, which is Option A.

Number plugging is not the best method to solve this question.

OA for this question is B, not A: the given inequality holds true in two ranges \(-1<x<0\) and \(x>1\) (see solution in my previous post). You can try values from this ranges to check. So x>2 is not always true.

x>2 will alwayz be true as it is a part of (-1,0)& (1,infinity) anyway, i realized my mistake. the answer will be B. thnx buddy!

No, that's not correct.

Given inequality holds true for \(-1<x<0\) and \(x>1\), so if \(x\) is in the range \(-1<x<0\) (for example if \(x=-0.5\)) or in the range \(1<x\leq{2}\) (for example if \(x=1.5\)) then \(x>2\) won't be true.

Very good question My attempt: given x/|x|<X and x is not equal to 0 option 1:if x > 2 let say x=3 then 3/3 = 1 which is < than 3 hence true. same can be stated about X= 2 hence must not be true.

option 2 : if -1<x<0 and 1<x<infinity. lets assume X is -0.5 then -1 < -.05 true. x=2 then 1< 2 true. this statement must be true for all values of X as per inequality given answer must be B
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If \(\frac{x}{|x|} \lt x\) , which of the following must be true about \(x\)? (\(x \ne 0\)) A. \(x\gt 2\) B. \(x \in (-1,0) \cup (1,\infty)\) C. \(|x| \lt 1\) D. \(|x| = 1\) E. \(|x|^2 \gt 1\)

Absolute value properties: Absolute value is always non-negative: \(|x|\geq{0}\) (not positive but non-negative, meaning that absolute value can equal to zero), so: When \(x\leq{0}\) then \(|x|=-x\) (note that in this case \(|x|=-negative=positive\)); When \(x\geq{0}\) then \(|x|=x\).

If \(\frac{x}{|x|} \lt x\) , which of the following must be true about \(x\)? (\(x \ne 0\)) A. \(x\gt 2\) B. \(x \in (-1,0) \cup (1,\infty)\) C. \(|x| \lt 1\) D. \(|x| = 1\) E. \(|x|^2 \gt 1\)

Absolute value properties: Absolute value is always non-negative: \(|x|\geq{0}\) (not positive but non-negative, meaning that absolute value can equal to zero), so: When \(x\leq{0}\) then \(|x|=-x\) (note that in this case \(|x|=-negative=positive\)); When \(x\geq{0}\) then \(|x|=x\).

A. \(x<0\) --> \(\frac{x}{-x}<x\) --> \(-1<x\) (not \(-1>x\) as you've written). But remember that we are considering the range \(x<0\), so \(-1<x<0\).

You could check that \(-1>x\) is not correct by plugging some number, say \(x=-10\) then \(\frac{x}{|x|}=-1>-10=x\).

Bunuel I got this question but the answer weren't the same as you have posted here. They had the original choices as posted in page 1. I got the correct solution but since the answer choices given were incorrect, I chose the wrong one.

Please do review this question on GC CATS.
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Bunuel I got this question but the answer weren't the same as you have posted here. They had the original choices as posted in page 1. I got the correct solution but since the answer choices given were incorrect, I chose the wrong one.