Is x^3 > x^2?
1. X > 0
2. X^2 > X
OA : C
Source: MGMAT Equations, 4th ed., pg. 187
Inequalities
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- anuprajan5
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Hi Troika,
The answer is C
Statement 1 - X>0
Since the question does not mention as to whether x is an integer, we could take fractions into the fray as well. We know that positive integers raised to a higher power will be greater than to the same integer raised to a lower power. But that does not hold for fractions. Examples:
2^3>2^2
But (1/2)^3<(1/2)^2
Insufficient
Statement 2 - x^2>x
This can hold for both negative and positive integers. But in the case of negative integers, this will not hold for the question.
2^2>2 and 2^3>2^2
(-2)^2>-2 but -2^3<-2^2
[spoiler]Combined, we know that the number is positive and that x^2>x^3. This holds true only for positive integers. Hence C[/spoiler]
The answer is C
Statement 1 - X>0
Since the question does not mention as to whether x is an integer, we could take fractions into the fray as well. We know that positive integers raised to a higher power will be greater than to the same integer raised to a lower power. But that does not hold for fractions. Examples:
2^3>2^2
But (1/2)^3<(1/2)^2
Insufficient
Statement 2 - x^2>x
This can hold for both negative and positive integers. But in the case of negative integers, this will not hold for the question.
2^2>2 and 2^3>2^2
(-2)^2>-2 but -2^3<-2^2
[spoiler]Combined, we know that the number is positive and that x^2>x^3. This holds true only for positive integers. Hence C[/spoiler]
Regards
Anup
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Anup
The only lines that matter - are the ones you make!
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