Is |x| < 1?

[sairamGmat]

. If x≠0, is |x| <1?
(1) x2<1
(2) |x| < 1/x


OA is D

I thought (1) will be true for all fractions between -1 and 1. So |x| < 1. - 1 is sufficient
In (2) , we can express the equation as x < 1/x => x2 < 1 - This is same as 1. => so |x| < 1
But, we can express this equation also as -x < 1/x which makes it x2 > 1. -> This will make |x| > 1.

So, I thought it is A. But not sure why is it OA. Help on this would be great...

[kvcpk]

. If x≠0, is |x| <1?
(1) x2<1
(2) |x| < 1/x


OA is D

I thought (1) will be true for all fractions between -1 and 1. So |x| < 1. - 1 is sufficient
In (2) , we can express the equation as x < 1/x => x2 < 1 - This is same as 1. => so |x| < 1
But, we can express this equation also as -x < 1/x which makes it x2 > 1. -> This will make |x| > 1.

So, I thought it is A. But not sure why is it OA. Help on this would be great...

You are right till this point:
-x < 1/x when x<0
now multiply both sides by x
We know that x is negative. So when we multiply an inequality with negative number, the inequality reverses.
For example. -2 < 1/2
now multiply both sides by -3, it should become
6>-3/2
But not 6<-3/2 .

So,
-x^2>1
x^2<-1 [By multiplying both sides by -1]
This is absurd.
Hence x is always greater than 0

Hope this helps!!

[anirban_lax]

.
But, we can express this equation also as -x < 1/x which makes it x2 > 1. -> This will make |x| > 1.

This is the point where error crept in. If -x < 1/x then
- x2 > 1 (multiplying both sides by the negative number x)
x2 < -1 (multiplying both sides by -1)

This would mean x2 is a negative number and hence x is an imaginary number. Now, in the GMAT all numbers are real unless specifically mentioned otherwise.
So, x<0 is invalid. Hence what we have is that x is a number greater than 0 yet less than it's reciprocal. A conclusion that can be drawn only for fractional numbers where numberator is less than the denominator. Hence x is fraction and is less than 1.

[roh00kan]

I didn't understand the statement 2 explanation.

|x| < 1/x

If x is +ve,
x < 1/x ==> x^2 < 1. This is okay. ---(a)


If x is -ve,
-x < 1/x ==> -x^2 < 1 {multiplied by x on both sides}.
OR x^2 > -1. This means, x^2 is > zero. --- (b) {squares cannot be negative}

Since, (a) says x^2 < 1 and (b) says x^2 > -1, we can say |x| < 1.

But how did you guys get
-x^2>1
x^2<-1

[gmatmachoman]

. If x≠0, is |x| <1?
(1) x2<1
(2) |x| < 1/x


OA is D

I thought (1) will be true for all fractions between -1 and 1. So |x| < 1. - 1 is sufficient
In (2) , we can express the equation as x < 1/x => x2 < 1 - This is same as 1. => so |x| < 1
But, we can express this equation also as -x < 1/x which makes it x2 > 1. -> This will make |x| > 1.

So, I thought it is A. But not sure why is it OA. Help on this would be great...

st 2: |x| < 1/x

Actually here we need to apply the basic numeric theroy concepts...else plug in numbers

case 1 : X = 2

|X| = 2

1/x = 1/2

|x| < 1/x.. Does our picked numbers satisfy the core condition?? Not really.

So again try X= 1/2

|X| = |1/2|= 1/2
1/X = 1/(1/2)
= 2


|x| < 1/x : 1/2 is LESS THAN 2 ...Does our picked numbers satisfy the core condition?? YES

So from our picked numebrs we can deduce that this condition |x| < 1/x can hold true only for 0<X<1

Now coming back to the main query is |X| < 1;

yes ofcourse since 0<X<1, all values of X will be less than 1

Pick D

[roh00kan]

gmatmachoman, thanks for your reply. I do agree with you.

But my question is different. I was asking how did previous 2 respondents, before you, get


-x^2>1
x^2<-1

and then they deduce that statement 2 is also sufficient and hence the answer is D. Am I missing something here?
Thanks

[kvcpk]

gmatmachoman, thanks for your reply. I do agree with you.

But my question is different. I was asking how did previous 2 respondents, before you, get


-x^2>1
x^2<-1

and then they deduce that statement 2 is also sufficient and hence the answer is D. Am I missing something here?
Thanks

Hi,

Let me come to your rescue

(2) |x| < 1/x

Two cases arise: 1)x>0, 2)x<0

Case1: x>0
When x>0, |x|=x
Hence x<1/x
x^2<1
x^2-1<0
This means that |x|<1

Case2: x<0
When x<0, |x|=-x
-x<1/x

Remember this thing. I have also stated the same on my above post.

Whenever an inequality is multiplied by a negative value, its sign reverses.
Example: 1<2 When multiplied by -1 on both sides,
it changes to -1>-2 and not -1<-2

With that in mind, I am trying to multiply both sides by x (which is negative)

Hence -x^2>1
Multiplying both sides by -1 again,
x^2<-1
this cannot be true. Because x^2 is positive.

hence this statement is sufficient.

Hope this helps!!

[roh00kan]

KVCPK, Thank you very much.

Well, I understood my mistake. I was considering the 'x' of 1/x as +ve and when I multiplied 'x' on both sides, I didn't reverse the sign as I was thinking that, 'x' is +ve.

[kvcpk]

KVCPK, Thank you very much.

Well, I understood my mistake. I was considering the 'x' of 1/x as +ve and when I multiplied 'x' on both sides, I didn't reverse the sign as I was thinking that, 'x' is +ve.

Glad it helpded