OR x>0 and |x|>|1|, [x>0, x>1] ==> x>1 Sufficient to answer No
st(2) |x|<|1| Not Sufficient
a
GmatKiss wrote:Is x negative?
1. X^3(1-x^2) < 0
2. x^2 - 1 < 0
NEGATIVE
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pemdas
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st(1) x<0 and 1>x^2 ~~ x<0 and |x|<|1|, [x<0, x>1]
OR x>0 and |x|>|1|, [x>0, x>1] ==> x>1 Sufficient to answer No
st(2) |x|<|1| Not Sufficient
a
OR x>0 and |x|>|1|, [x>0, x>1] ==> x>1 Sufficient to answer No
st(2) |x|<|1| Not Sufficient
a
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Statement 1: x³(1 - x²) < 0GmatKiss wrote:Is x negative?
1. X^3(1-x^2) < 0
2. x^2 - 1 < 0
--> x³(1 - x)(1 + x) < 0
--> either -1 < x < 0 or x > 1
Not sufficient
Statement 1: (x² - 1) < 0
--> x² < 1
--> -1 < x < 1
Not sufficient
1 & Together: -1 < x < 0
Hence, x is negative.
Sufficient
The correct answer is C.
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Statement 1: x³(1-x²) < 0.GmatKiss wrote:Is x negative?
1. X^3(1-x^2) < 0
2. x^2 - 1 < 0
x³(1+x)(1-x) < 0.
The critical points are x=0, x=-1, x=1.
These are the only values where x³(1+x)(1-x) = 0.
When x is any other value, x³(1+x)(1-x) < 0 or x³(1+x)(1-x) > 0.
To determine the range of x, test one value to the left and right of each critical point.
x^3(1-x^2) < 0 can be rephrased as x³ < x^5.
Plug x < -1 into x³ < x^5:
Let x = -2.
(-2)³ < (-2)^5
-8 < -32.
Doesn't work.
x<-1 is NOT part of the range.
Plug -1<x< 0 into x³ < x^5:
Let x = -1/2.
(-1/2)³ < (-1/2)^5.
-1/8 < -1/32.
This works.
-1<x<0 is part of the range.
Plug 0<x<1 into x³ < x^5:
Let x = 1/2.
(1/2)³ < (1/2)^5
1/8 < 1/32.
Doesn't work.
0<x<1 is NOT part of the range.
Plug x > 1 into x³ < x^5:
Let x = 2
2³ < 2^5
8 < 32.
This works.
x>1 is part of the range.
Two ranges work in statement 1:
-1<x<0 and x>1.
Since x can be negative or positive, INSUFFICIENT.
Statement 2: x²-1<0.
(x+1)(x-1) < 0.
The critical points are x = -1 and x = 1.
These are the only values where x²-1 = 0.
When x is any other value, x²-1 < 0 or x²-1 > 0.
To determine the range of x, test one value to the left and right of each critical point.
x²-1<0 can be rephrased as x²<1.
Plug x < -1 into x²<1:
Let x = -2.
(-2)² < 1.
4 < 1.
Doesn't work.
Plug -1 < x < 1 into x²<1:
Let x = 0.
0² < 1.
This works.
-1<x<1 is part of the range.
Plug x > 1 into x²<1:
Let x = 2.
2² < 1.
4 < 1.
Doesn't work.
The only range that works is -1 < x < 1.
Since x can be negative, 0, or positive, INSUFFICIENT.
Statements 1 and 2 combined:
Ranges that satisfy statement 1: -1 < x < 0 or x>1.
The only range that satisfies statement 2: -1 < x < 1.
The only range that satisfies BOTH statements is -1<x<0.
Since x must be negative, SUFFICIENT.
The correct answer is C.
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