Is x negative? (1) x^3(1 - x^2) < 0 (2) x^2 - 1 < 0

Is x negative?


(1) \(x^3(1-x^2)<0\), two cases (one of the multiples negative, another positive):

A. \(x^3>0\) and \(1-x^2<0\);

\(x^3>0\) --> \(x>0\)
----------------0----------------

\(1-x^2<0\) --> \(x<-1\) or \(x>1\)
-----(-1)------0------(1)-------

Hence the range for A is \(x>1\).

B. \(x^3<0\) and \(1-x^2>0\)

\(x^3<0\) --> \(x<0\)
---------------0------------

\(1-x^2>0\) --> \(-1<x<1\)
-----(-1)------0------(1)-----

Hence the range for B is \(-1<x<0\).

Two ranges: \(x>1\) OR \(-1<x<0\). Not sufficient.


(2) \(x^2-1 < 0\) --> \(-1<x<1\). Not sufficient.


(1)+(2) Intersection of the ranges from (1) and (2) is \(-1<x<0\), hence \(x<0\) is true. Sufficient.

Answer: C.
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IMO C:

A) X^2(1-X^2) <0 so X^3<0 or X^2>1
x could be positive or negative

B) X^2-1<0
(X-1)(X+1)<0 so X<1 or X<-1
x could be positive or negative

Combined X is negative for both statements to hold true.

OA?

IMO C.

1, is equivalent to x(x-1)(x+1) >0 ( x^2 is always>0 and i have inverted the sign)

this gives x>1 and -1<x<0 , thus not sufficient

2, gives -1< x<1 not sufficient.

if you combine the result we get -1<x<0 , thus x is negative and thus C.

Another way, since we know both are not sufficient individually.

2nd equation gives x^2 -1 <0 => 1- x^2 >0
now in first equation, if its -ve then x^3 must be -ve....thus both taken together sufficient.
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Alternative approach by taking numbers.
Statement1: \(x^3(1-x^2) < 0\)
From this we can say \(x <> 0\)
and \(x <>+1 or -1\).

Consider x=2 Then the equation results in \(8*-3 <0\). Satisfy the equation.
Consider x= 1/2 then the equation results in \(1/8*(3/4) >0\). Does not satisfy the equation.

Consider x=-2 Then the equation results in \(-8*-3 >0\). Does not satisfy the equation.
Consider x= -1/2 then the equation results in \(-1/8*(3/4) <0\).Satisfy the equation.

This statement is not sufficient to conclude.

Statement2: \(x^2-1 <0\)
From the statement we can say that \(x <>=1 or -1\)
Consider x=2 then the equation results in \(3 >0\). Does not satisfy the equation.
Consider x=1/2 then the equation results in \(-3/4 <0\). Satisfy the equation.

Consider x=-2 then the equation results in \(3 >0\). Does not satisfy the equation.
Consider x=-1/2 then the equation results in \(-3/4 <0\). Satisfy the equation.

This statement alone is not sufficient.

Combining both the statements when x is -ve that is -1/2 , it satisfies both the equations.

Ans is C.

1. x^3(1-x^2) <0
x^3- x^5<0
x^3 < x^5 , which mean that x is positive no negative. ( sufficinet).


2. x^2<1

-1 < x < 1

which mean that we do not know exactally wheather it is positive or negative. ( not sufficient).


the answer is A

Is \(x<0\)?

(1) \(x^3(1-x^2) < 0\)
Two cases:
A. \(x>0\) and \(1-x^2<0\), which means \(x<-1\) or \(x>1\) --> \(x>1\);
B. \(x<0\) and \(1-x^2>0\), which means \(-1<x<1\) --> \(-1<x<0\).

So \(x^3(1-x^2) < 0\) holds true for two ranges of \(x\): \(x>1\) and \(-1<x<0\). If \(x\) is in the first range answer to the question is NO, but if \(x\) is in the second range answer to the question is YES. Two different answers. Not sufficient.

(2) \(x^2-1<0\) --> \(-1<x<1\). x can be positive as well as negative. Not sufficient.

(1)+(2) \(x>1\) or \(-1<x<0\) AND \(-1<x<1\) --> intersection of the ranges from (1) and (2) is \(-1<x<0\). \(x\) is negative. Sufficient.

Answer: C.



Red part is not correct. \(x^3 < x^5\) does not hold true for all positive \(x-es\). For example \(x=0.5\) --> \(x^3>x^5\). Also \(x\) can be negative in the range \(-1<x<0\) and \(x^3 < x^5\) will hold true.
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Question: Is x negative?

Please check the attachment for detailed solution.

This is how GMATINSIGHT teaches Data sufficiency

Answer: Option C

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My reasoning:

Statement 1: either \(x^3\) is negative or \((1 - x^2)\) is negative. They are opposite signs. However we can have a positive x and a negative x in this case.

Testing values of 2 and \(\frac{-1}{3}\) will prove this.

Statement 2: \(x^2 - 1 < 0\) -> \(x^2 < 1\) so \(x > -1 & x < 1\) Again inconclusive because we can have any positive or negative fraction.

Combining the two statements only "negative fractions" are in our overlapping set. The answer is C. x MUST be less than 0

Hi chetan2u, Bunuel, Gladiator59

I am not able to figure out how we are able to find solution combining both the statement for above question, Could you please assist.

The question is


Let us take both combined..

Statement II tells us \(x^2 - 1 < 0\), so when we multiply this with -1, the inequality sign will change
=>\(-1(x^2 - 1) > 0*(-1).....1-x^2>0\) We did this to get the term of statement I..

Now take statement I..
\(x^3(1 - x^2) < 0\) ... Here we see from statement II that \(1-x^2>0\), so \(x^3(1 - x^2) < 0\) => \(x^3(positive..value) < 0\) , thus \(x^3<0...x<0\)
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Is x negative?


(1) x^3(1−x^2)<0

(2) x^2−1<0

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IMO, it is C.


The question is x < 0?
(1) divide both RHS and LHS by x^2.
then you have, x (1-x^2) <0.
now, take "-1" out from the parenthesis. -x(x^2 - 1) < 0.
Divide both RHS and LHS by -1.
then you have x (x^2 - 1) > 0.
In other words, x ( x+1) (x-1) >0.
so -1<x<0 and x>1
insufficient.

(2) we can say ( x+1) (x-1) <0
so -1<x<1
insufficient.

(1) + (2) together
-1<x<0.
so x is negative.
sufficient to answer whether x is negative.

Graphical solution:

1) \(x^3(1 - x^2) < 0\)
Case I: \(x^3 < 0\) and \((1 - x^2) > 0\)
x < 0 and \(x^2 < 1\)
x < 0 and - 1 < x < 1
So, - 1 < x < 0 YES

Case II: \(x^3 > 0\) and \((1 - x^2) < 0\)
x > 0 and \(x^2 > 1\)
x > 0 and x > 1 and x < -1
So, x > 1 YES

INSUFFICIENT.

2) \(x^2 - 1 < 0\)
\(x^2 < 1\)
- 1 < x < 1
YES and NO both cases.

INSUFFICIENT.

Together 1 and 2
Common range is -1 < x < 0

SUFFICIENT.

Answer C.
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Is x negative?
Is \(x < 0\)?

(1) \(x^3(1 - x^2) < 0\)

Either x^3 or (1 - x^2) is negative.

If \(x^3 < 0\) then

\(1 - x^2 > 0\)
\(x^2 < 1\)
\(-1 < x < 1\)

If \(x^3 > 0\) then
\(1 - x^2 < 0\)
\(x^2 - 1 > 0\)
\((X+1)(X-1) > 0\)
\(X > 1\) and \(x < -1\)

INSUFFICIENT.

(2) \(x^2 - 1 < 0\)
\((x+1)(X-1) < 0\)
\(-1 < x < 1 \)

INSUFFICIENT.

(1&2) Combined, the range is \(-1 < x < 0\).

Thus, \(x < 0\).

SUFFICIENT.
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Could someone explain how can we transform x^2-1<0 to -1<x<1. I mean I understand it logically but algebraically we have (x+1)(x-1)<0 => x<-1 or x<1

It's a matter of rearranging

x^2 - 1 < 0
x^2 < 1
\sqrt{x^2} < \sqrt{1}
| x | < 1 <-----MEMORIZE this...anytime you see the square root of a variable raised to an even exponent, you end up with the absolute value of that variable

Hence, -1 < x < 1.

This might be a silly question, but suppose after combining both statements we find that there are no areas of overlap. Then the answer would be E?
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Is X < 0 ?


(S1) (x)^3 * (1 - (x)^2) < 0


Case 1: (x)^3 < 0 ------and------- 1 - (x)^2 > 0

X < 0 ------ and -------(x)^2 < 1

X < 0 ------and------- (-)1 < X < +1


Summary of Case 1: if X < 0 -------- (-)1 < X < 0



Case 2: (x)^3 > 0 ------ and -------- 1 - (x)^2 < 0

X > 0 ----- and ------ (X)^2 > 1


Summary of Case 2: if X > 0 ------- X > +1

X can be Negative or Positive

S1 NOT Sufficient alone


(2)
(x)^2 < 1

[X] < 1

-1 < X < +1

X can be Negative or Positive
S2 NOT Sufficient Alone


Together:

If X < 0:

Statement 1 says: -1 < X < 0
and
Statement 2 says: -1 < X < 0


If X > 0

Statement 1 says: X > +1
and
Statement 2 says: 0 < X < +1

if X is a Positive Value, it can NOT Satisfy S1 and S2 TOGETHER at the Same Time


therefore, X < 0

and Definite YES: X is Negative

(C) Together Sufficient

Bunuel chetan2u VeritasKarishma

Could you please explain why the below solution is incorrect?

When x is between 0 and -1, say x is -1/2.
\(\frac{-1}{2}^3=\frac{-1}{8}\)
\(\frac{-1}{2}^5=\frac{-1}{32}\)
\(\frac{-1}{32}>\frac{-1}{8}\)

So x can be negative too.
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\(x^3 < x^5\)

does not imply that x is positive. This is satisfied for x = 2 as well as x = -1/2.
Note that the number properties vary over 4 basic intervals of the number line

x < -1
-1 < x< 0
0 < x < 1
and x > 1

Also, -1, 0 and 1 behave differently. It is best to use

\(x^3 - x^5 < 0\)
\(x^5 - x^3 > 0\)
\(x^3 (x^2 - 1) > 0\)
\(x^3 (x + 1)(x - 1) > 0\)

Transition points are 0, -1, 1.
x will be positive when x > 1 or -1 < x < 0.
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