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12 Feb 2012, 15:07
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
50% (01:34) correct
50%
(01:47)
wrong
based on 92
sessions
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Is x between 0 and 1
(1) x^2 > x^3
(2) -x > x^3
(1) x^2 > x^3
(2) -x > x^3
I solved this as below
x^3 - x^2 < 0 or x^2( x - 1 ) < 0 roots will be x = 0 and x = 1 or
0 < x < 1 why is this wrong ?
x^3 - x^2 < 0 or x^2( x - 1 ) < 0 roots will be x = 0 and x = 1 or
0 < x < 1 why is this wrong ?
12 Feb 2012, 15:46
Kudos
Bookmarks
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 = B
The 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
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 = B
The 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
13 Feb 2012, 06:40
Kudos
Bookmarks
rxs0005
Is x between 0 and 1
1. x^2 > x^3
2. -x > x^3
I solved this as below
x^3 - x^2 < 0 or x^2( x - 1 ) < 0 roots will be x = 0 and x = 1 or
0 < x < 1 why is this wrong ?
the OA is B
1. x^2 > x^3
2. -x > x^3
I solved this as below
x^3 - x^2 < 0 or x^2( x - 1 ) < 0 roots will be x = 0 and x = 1 or
0 < x < 1 why is this wrong ?
the OA is B
Is x between 0 and 1
(1) x^2 > x^3 --> x^2*(x-1)<0, since x^2 here must be positive, then x-1 must be negative: x-1<0 --> x<1. Not sufficient.
(2) -x>x^3 --> x(x^2+1)<0 --> since x^2+1 is positive then x must be negative: x<0, so the answer to the question is NO. Sufficient.
Answer: B.
Hope it's clear.
