Hi, there. I'm happy to help with this.
Prompt:
Is x between 0 and 1?
Statement #1: x^2 > x^3
Consider four categories of numbers
(a) positive numbers bigger than one
(b) positive numbers between one and zero
(c) negative numbers between 0 and -1
(d) negative numbers less than -1 (i.e. with absolute value greater than 1)
This is always a good list to keep in mind, especially on DS questions on exponents.
Consider the four categories:
(a) x = 2 ---> 4 < 8, statement #1 is not true
(b) x = 0.5 ---> 0.25 > 0.125, statement #1 is true
(c) x = -0.5 ---> +0.25 > -0.125, statement #1 is true
(d) x = -2 ---> +4 > -8, statement #1 is true
So, statement #1 allows for several categories, not just between 0 and -1. Statement #1 is
insufficient.
Statement #2: -x > x^3
If x is positive, this will never be true, because -x would be negative, x^3 would be positive, and any positive is greater than any negative.
If x is negative, this will always be true, because -x would be positive, x^3 would be negative, and any positive is greater than any negative.
If x is negative, it's not between 0 and 1. This allows us to give a conclusive answer to the prompt. Statement #2, by itself, is
sufficient.
Answer =
BThe problem with your algebraic method -- you factored correctly, and arrived at:
(x^2)(x-1) < 0
but this is a
cubic inequality. While you do need to find the roots, it's not enough just to say that the solution exists between them. The roots delimit regions of the number line, and we need to test the inequality in each region. Thus,
Region I = less than 0
Region II = between 0 and 1
Region III = greater than 1
Upon testing values, we find that numbers in both Region I and Region II satisfy this inequality. Thus, this inequality by itself does not guarantee that the numbers are or are not between 0 and 1.
Does that make sense? Please let me know if you have questions on anything I've said here.
Mike
_________________
Mike McGarry
Magoosh Test Prep
close
60% (01:33) correct
