tracyyahoo wrote:1.
Is x<0?
(1) x^3<x^2
(2) x^3<x^4
Statement 1:
x³ - x² < 0.
x²(x-1) < 0.
The critical points are x=0 and x=1.
These are the only values of x where x³ - x² = 0.
When x is any other value, x³ - x² < 0 or x³ - x² > 0.
To determine the range of x, test one value to the left and right of each critical point.
x<0:
Plug x=-1 into x³ - x² < 0:
(-1)³ - (-1)² < 0.
-2<0.
This works.
x<0 is part of the range.
0<x<1:
Plug x= 1/2 into x³ - x² < 0:
(1/2)³ - (1/2)² < 0.
-1/8 < 0.
This works.
O<x<1 is part of the range.
x>1:
Plug x=2 into x³ - x² < 0:
(2)³ - (2)² < 0.
6<0.
Doesn't work.
x>1 is not part of the range.
Two ranges satisfy statement 1: x<0 and 0<x<1.
Thus, we cannot determine whether x<0.
Insufficient.
Statement 2:
x^3 - x^4 < 0.
x³(x-1) < 0.
The critical points are x=0 and x=1.
To determine the range of x, test one value to the left and right of each critical point.
x<0:
Plug x=-1 into x³ - x^4 < 0:
(-1)³ - (-1)^4 < 0.
-2<0.
This works.
x<0 is part of the range.
0<x<1:
Plug x= 1/2 into x³ - x^4 < 0:
(1/2)³ - (1/2)^4 < 0.
1/16 < 0.
Doesn't work.
0<x<1 is not part of the range.
x>1:
Plug x=2 into x³ - x^4 < 0:
(2)³ - (2)^4 < 0.
-8<0.
This works.
x>1 is part of the range.
Two ranges satisfy statement 2: x<0 and x>1.
Thus, we cannot determine whether x<0.
Insufficient.
Statements 1 and 2 combined:
Only one range satisfies BOTH statements: x<0.
Sufficient.
The correct answer is
C.
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