Is integer x positive?

You can use inequalities to solve this question or you can also solve it easily if you have a good understanding of the relation between x, x^2 and x^3 in various ranges. In fact, add \(\sqrt{x}\) to the list as well (though it is not needed in this question).

Method 1: Inequalities

(1) x > x^3
x^3 - x < 0
x(x - 1)(x + 1) < 0
-1, 0 and 1 are the transition points. Inequality will hold when 0 < x <1 or x < -1. So x cannot be integer and positive both. Sufficient.

(2) x < x^2
x^2 - x > 0
x(x - 1) > 0
0 and 1 are the transition points. Inequality will hold when x > 1 or x < 0. x can be an integer and either positive or negative. Not sufficient.

Answer (A)

Method 2: It is easy to see that A is enough if you understand the relation between x, x^2 and x^3. Look at the number lines below.



The green regions show the areas for which the relations hold. You can see that x^3 < x only for negative integer values (and some positive but non integer values) of x. Hence statement 1 is enough.
Answer (A)
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My attempt:

Is integer X positive?

1. X > X^3

A number is greater than number raised to an odd powers implies that the number is an negative integer since a number raised to an odd power would retain its sign.

Ex:

-2 > (-2)^3

Hence X is negative. Sufficient

2. X < X^2

A number raised to an even power would mask its sign. Hence X^2 would always be positive.

Ex :1
-2 < (-2)^2

Ex :2
3 < 3^2

Hence X can be either positive or negative.

Insufficient

Answer is A is alone sufficient to solve this problem.

I'd say C:

The question is: is X>0 ?

1) X > X^3

X can be negative: If X=5, -5 > -75 is true
X can be positive: If X=0,3 0,3 > 0.027 is true

Hence, not sufficient

2) X < X^2

X can be negative: If X=-5, -5<25 is true
X can be positive: If X=2, 2<4 is true

Hence, not sufficient

Together:
The first answer brings two options: either X<0 or 0<X<1
The second answer brings two options: either X<0 or 1<X

X cannot be both greater than 1 and smaller than one, hence it is negative. Answer C

I thought a similar thing at first, however on closer inspection I noticed that the original question mentions that X is an integer - therefore we can eliminate the cases where -1<X<1

What about 1/2 ?

1/2 > 1/8

the answer should be E

stmt 1 ==> x>-1 and 0<X<1

stmt 2 ==> X<0 and X>1

The questions says that X is an integer. No need to check for x=1/2.

Answer is A.

A)\(x > x^3\) in case 0 < x < 1 or if x < (-1)
but given x is integer so x is -ve number
sufficient
B) \(x < x^2\)
x<0 or
if x>1,
so x could be negative as well as positive. Not sufficient

so A

I got it wrong.. i missed the word "integer" in the question.

(1) Sufficient. \(x>x^3\) is the same as \(x^3-x<0\) or \(x(x^2-1)<0\).
\(x^2-1>0\) for any integer \(x\) except -1, 0 1 which are not appropriate for us. So \(x<0\)

(2) Insufficient. It could be -1 or 2 as well.

The correct answer is A.

Answer A cannot be right.

If you put in 1/2 or -2 the result will be the same. but different answers to the question!

Sorry, my mistake oversaw the integer!

Hi Karishma,

Can you please explain how is x <1 or x > 1
My answer came x > 0 or x > 1.
How to solve that inequality??

Hi All,

This question can be approached in a couple of different ways, but you might find that using Number Properties and TESTing VALUES is a pretty fast way to get to the answer.

We're told that X is an INTEGER. We're asked if it is POSITIVE. This is a YES/NO question.

Fact 1: X > X^3

I'm going to TEST VALUES to prove a pattern...

IF....
X = 1
X is NOT > X^3....so X CANNOT be 1

IF...
X = 2
X is NOT > X^3....so X CANNOT be 2

IF....
X=3
X is NOT > X^3....so X CANNOT BE 3

As X increases, X^3 increases even more, so X CANNOT be any positive integer. Since the question asks if X is POSITIVE, we have the answer: NO, X is NEVER positive.
Fact 1 is SUFFICIENT

Fact 2: X < X^2

Here, we can TEST VALUES to prove that there are multiple answers....

IF....
X = 2
X IS < X^2 and the answer to the question is YES.

IF....
X = -1
X is < X^2 and the answer to the question is NO.
Fact 2 is INSUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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When we solve x(x - 1)(x + 1) < 0,
we will get that inequality will hold when 0 < x <1 or x < -1.
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1) X>X3
dont we need to check inequality for four
a) X<-1 - true
b) -1<x<0 - true
c) 0<x<1 true
d) x>1 - false

Official Solution:

Is integer x positive?

(1) \(x \gt x^3\).

\(x^3-x \lt 0\);

\(x(x-1)(x+1) \lt 0\);

\(x \lt -1\) or \(0 \lt x \lt 1\).

Since we are told that \(x\) is an integer and there is no integer in the range \(0 \lt x \lt 1\) then \(x \lt -1\), so \(x\) is a negative integer. Sufficient.

(2) \(x \lt x^2\).

\(x^2-x \gt 0\);

\(x(x-1) \gt 0\);

\(x \lt 0\) or \(x \gt 1\), so integer \(x\) could be negative as well as positive. Not sufficient.


Answer: A
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Is integer x positive?

Stat1: x > x^3
or, x*(1-x^2) > 0;
Case1- x >0 then (1-x^2) > 0 or, -1 < x < 1. So, 0 < x < 1. But x = an Integer, so no value of x satisfies in this case.
Case2- x <0 then (1-x^2) < 0 or, x < -1 or x > 1. So, x < -1. Sufficient

Stat2: x < x^2[/quote]
x* (1-x) <0
Case1- x >0 then (1-x) > 0 or, x < 1. So, 0 < x < 1. But x = Integer, so no value of x in this case.
Case2- x <0 then (1-x) < 0 or, x > 1. So, x < 0 or x >1. Not Sufficient

So, I think A.

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