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

If \(x\neq{0}\), is \(|x| <1\)?

Is \(|x| <1\)?

Is -\(1<x<1\) ? (\(x\neq{0}\))

(1) \(x^2<1\) --> \(-1<x<1\). Sufficient.

(2) \(|x| < \frac{1}{x}\)

Since LHS (|x|) is an absolute value, which is always non-negative, then RHS (1/x), must be positive (as \(|x| < \frac{1}{x}\)), so \(\frac{1}{x}>0\) --> \(x>0\).

Now, if \(x>0\) then \(|x|=x\) and we have: \(x<\frac{1}{x}\) --> since \(x>0\) then we can safely multiply both parts by it: \(x^2<1\) --> \(-1<x<1\), but as \(x>0\), the final range is \(0<x<1\). Sufficient.

Answer D.
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If x≠0, is |x| < 1 ?

is x<1
OR
is -x<1
x>-1

Is -1<x<1?

(1) x^2 < 1
x^2<1
|x|<1

This tells us exactly what the stem looks for.
SUFFICIENT

(2) |x| < 1/x
is x < 1/x
OR
is -x < 1/x
is x > -1/x
SO
-1/x < x < 1/x
-1 < x^2 < 1
SUFFICIENT

(I am going to use Bunuel's method only because I think it makes more sense and I believe mine is wrong anyways)

|x|<1/x
if |x|<1/x then 1/x MUST be positive as it is greater than an absolute value. If 1/x is positive then x must also be positive and therefore |x|<1/x is actually equal to x<1/x. Because we know that x is positive we can multiply both sides by x to simplify.

(x)* x < 1/x *(x)
x^2 < 1
|x|<1
x<1 or x>-1
-1<x<1

This tells us exactly what the stem is looking for
SUFFICIENT
(D)

(also, could someone explain to me how to solve by the method I originally chose in #2...the one where I get the positive and negative case for |X|)

Thanks!

My answer was A...
Explanation->
|x|<1 means x should be between -1 and 1.

from (1)
x2-1<0 => x between -1 and 1. sufficient

from (2)
if x +ve then same as 1 and sufficient
if x -ve then |x|=-x
-x2<1 or x2 > -1 not sufficient

so and A

Please correct me if wrong

Merging similar topics.



\(x\) cannot be negative. Refer to the solution above.

Also if \(x<0\) then we have \(-x<\frac{1}{x}\) and now if we cross multiply by negative \(x\) then we should flip the sign: \(-x^2>1\) --> \(x^2+1<0\) which cannot be true for any real value of \(x\) (the sum of two positive value cannot be less than zero).
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Given question stem asks if |x|<1------> Is -1<x<1

from St 1 we have x^2<1 ------> -1<x<1 So Sufficient

from St2 we have

|x|<1/x

Notice that |x| is a positive value and for any Integer value |x|> 1/x -----This implies X is a fraction. In order to satisfy the above equation let us take some fractional value of x and check what happens to the above equation

x= -1/2 so we have 1/2<-2 ------No
x=3/4 so we have 3/4< 4/3 ------Yes
x=4/3 so we have 4/3 <3/4 ------- no

We see that when fraction is between 0<x<1 then the above equation holds true and hence x is between -1 and 1
Ans should be D....

Bunuel's solution is superb. Saves time
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If x≠0, is |x| < 1 ?

(1) x^2 < 1
x<1, x>-1
-1<x<1

Valid X is a number that is greater than -1 (i.e. -3/4, -1/2) and less than 1 (i.e. 1/2, 3/4) either way the absolute value of any of those numbers will be less than 1.
SUFFICIENT

(2) |x| < 1/x
Multiply both sides by x
x^2<1
Same solution as above.
SUFFICIENT

(D)

Is it valid to multiply the absolute value of x on one side (like in number two) to simplify as long as x is positive?

As, since from |x| < 1/x we concluded that x is positive, then yes we can do that: x < 1/x --> x^2 < 1.

Hope it helps.
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Hi Mahmud,

|x| <1 can be written as -1<x<1 because 1 is a constant BUT

|x| < \(\frac{1}{x}\) cannot be written as -1/x<x<1/x because 1/x is a variable

Solving,

|x| <\(\frac{1}{x}\)

RHS has to be greater than 0 (As LHS can only be +ve or 0)
=> \(\frac{1}{x}\) > 0
=> x>0 (x cannot be -ve or 0 ,
Because, if x is -ve then \(\frac{1}{x}\) is -ve
if x = 0 then\(\frac{1}{x}\) is not defined)

Rgds,
Rajat

Hi Bunuel!

A question here.

How is St 2 sufficient? You say it yourself that 2) implies that 0<x<1 so that is just part of the range of the question stem ( is -1<x<1 ?) So statement two doesn't cover the negative area of the range. How can it be considered to be sufficient?

Do I make myself clear btw?

The question asks: whether x is between -1 and 1. (2) says that x is between 0 and 1, so the answer to the question is YES.
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If \(x\neq{0}\), is \(|x|\)< 1?
This means -1<x<1, where \(x\neq{0}\)

(1) \(x^2\) < 1
\(x^2<1\) means |x|<1, since only fractions with absolute value less than 1 will have their square less than 1.
Sufficient

(2) \(|x|\) < \(1/x\)
Since |x| >0, we can write 1/|x|x>1
Since |x| and 1 are positive, x also must be positive for above to be true.
And since X is also positive, |X|=X, and X<1/X can be written as x^2<1
Therefore X is between 0 and 1
Sufficient

D
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Can you please elaborate on how a "partial" range in 2) (0 < x < 1) satisfies the full range that is required by the original prompt? I would think that the ranges have to be exact same, i.e. 1) provides a range that is exact same as compared to that in the prompt.

Thanks.

Any x, which is between 0 and 1, will obviously fall into the range between -1 and 1, so we can answer the question with an YES: x IS between -1 and 1.
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This is what I did for statement 2:

|x|<1/x
As |x| will always be non negative, so the equation becomes:
X<1/x
Multiplying both sides by x
X^2<1
X^2-1<0 which is then to be worked out as the same way for st 1.

Experts please let me know is this method correct? Bunuel chetan2u VeritasKarishma
Please help

Posted from my mobile device

Yes Shef08, you can do this here because \(\frac{1}{x}>|x|\) means \(\frac{1}{x}\geq{0}\), which in turn means \(x\geq{0}\), so you can cross multiply..

exc4libur , as we can see BOTH sides are non-negative, we can square both sides to get our answer ..
\(\frac{1}{x}>|x|\)......\(\frac{1}{x^2}>|x|^2....x^4<1\)
Now x>0, so 0<x<1
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If we write |x| < 1/x as x < 1/x, then we are assuming that x is positive because |x| = x only when x >= 0.

x < 1/x
I cannot multiply both sides by x without knowing the sign of x. Here x could be positive or negative e.g.
when x = 1/2, x < 1/x
when x = -2, x < 1/x

So I won't multiply both sides by x. I take 1/x to the other side.

x - 1/x < 0

(x^2 - 1)/x < 0

So 0 < x < 1 or x < -1
But x must be positive so only 0 < x < 1 satisfies.

Now try putting |x| = -x when x is negative.
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A Masterpeice from OG

By working on the Question stem, we will have -1<x<1,

S1. x^2 < 1, it means -1<x<1. Sufficient
S2. |x| < 1/x, it means x has to be positive, because | | is always positive and it is less than a value. So x has to be positive.
Now as x is positive, and the reciprocal is greater than a number, it means the number has to be between 0 & 1. Sufficient.

Hence D, as both statement alone is sufficient

[quote="KarishmaB"]

KarishmaB

Hi Karishma 🙏

please advise in x-1/x<0

(X^2-1)/x<0
(x-1)(x+1)/x<0
Now x cannot be zero. As denominator can not zero

But how do we get

0 < x < 1 or x < -1

I understand that x must be positive

Posted from my mobile device

Check out this post:
https://gmatclub.com/forum/inequalities ... 91482.html

Let me know if you have any doubts in it.
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