waggrave wrote:Is a > 0 ?
(1) a^3 - a < 0
(2) 1 - a² > 0
I received a PM requesting that I show how this DS can be solved by determining the
critical points.
Statement 1: a³ - a < 0.
a(a+1)(a-1) < 0.
The critical points are a=0. a=-1, a=1.
These are the only values where a(a+1)(a-1) = 0.
When a is any other value, a(a+1)(a-1) < 0 or a(a+1)(a-1) > 0.
To determine the range of a, test one value to the left and right of each critical point.
Plug a < -1 into a³ - a < 0:
Let a = -2.
(-2)³ - (-2) <0.
-6 < 0.
This works.
Plug -1 < a < 0 into a³ - a < 0:
Let a = -1/2.
(-1/2)³ -(-1/2) < 0.
3/8 < 0.
Doesn't work.
Plug 0 < a < 1 into a³ - a < 0:
Let a = 1/2.
(1/2)³ - 1/2 < 0.
-3/8 < 0.
This works.
Plug a > 1 into a³ - a < 0:
Let a = 2
(2)³ - 2 < 0.
6 < 0.
Doesn't work.
Two ranges work in statement 1:
a < -1.
0 < a < 1.
Since a can be negative or positive, insufficient.
Statement 2: 1 - a² > 0
1 - a² > 0
(1+a)(1-a) > 0.
The critical points are a = -1 and a = 1.
These are the only values where 1 - a² = 0.
When a is any other value, 1 - a² < 0 or 1 - a² > 0.
To determine the range of a, test one value to the left and right of each critical point.
Plug a < -1 into 1 - a² > 0:
Let a = -2.
1 - (-2)² > 0.
-3 > 0.
Doesn't work.
Plug -1 < a < 1 into 1 - a² > 0:
Let a = 0.
1 - 0² > 0.
1 > 0.
This works.
Plug a > 1 into 1 - a² > 0:
Let a = 2.
1 - 2² > 0.
-3 > 0.
Doesn't work.
The only range that works is -1 < a < 1.
Since a can be negative or positive, insufficient.
Statements 1 and 2 combined:
Ranges that satisfy statement 1: 0 < a < 1 or a < -1.
Range that satisfies statement 2: -1 < a < 1.
The only range that satisfies both statements is 0 < a < 1.
Since a must be positive, sufficient.
The correct answer is
C.
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