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DS question

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Source: — Data Sufficiency |

by mschling52 » Mon Jun 11, 2007 1:33 pm
One way to do this would be to find the zeroes of the functions and plot them on a number line, then evaluate the function for a value between each zero.

For (1), we have zeroes at x = -1, 0, and 1. If we pick something less than -1, we get a positive value since x^3 and 1-x^2 will both be negative. Then pick something between -1 and 0, then between 0 and 1, and finally something greater than 1. Continuing this process the number line will look like

-------(-1)-------(0)-------(+1)-------
__+_______-_______+______-___ (sign of (x^3)(1-x^2))

So, we know that a negative value of the function can be the result of x being -1<x<0 or x>1 and we can't say if x is negative.

From (2), we can follow a similar process to get a number line like

-------(-1)-----------------(+1)-------
__+___________-___________+__ (sign of (x^2)-1)

So we know that x satisfies -1<x<1, but can't say if x is negative.

Combining (1) and (2), we know that x is between -1 and 1 from (2). Lining this up with the 2 scenarios that satisfy (1), we can get that -1<x<0, so x is negative.
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Re: DS question

by jayhawk2001 » Mon Jun 11, 2007 4:04 pm
dunkin77 wrote:Is X negative?

(1) X^3(1-X^2)<0
(2) X^2-1<0
Posting an alternative soln...

2 tells us x^2-1 < 0, so 1-x^2 > 0.

If x^3 * (1-x^2) < 0 and 1-x^2 > 0, then x^3 is negative and so
x has to be negative.
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