xiamsamx wrote:Is a>0?

(1) a^3 - a < 0
(2) 1 - a^2 > 0


I know that you can solve theoretically but I wanted to see where I went wrong when I tried to solve algebraically.

Here is what I did:

1) a^3 - a < 0 can be rewritten as a*(a^2 - 1) < 0. I know one solution is a <0.
Looking at (a^2 - 1) < 0, I then rewrote that as (a+1)(a-1) < 0.
My second set of solutions then is a < -1 and a < 1.
Statement 1 is insufficient.

2) 1 - a^2 > 0 can be rewritten as (1+a)(1-a) >0. Your solutions are then a > -1 and a < 1.
Statement 2 is also insufficient.

Looking at both statements together, you still get a range of -1 < a < 1, so I picked answer choice E.

The correct answer is C.

1) a^3 - a < 0 can be rewritten as a*(a^2 - 1) < 0. I know one solution is a <0.
Looking at (a^2 - 1) < 0, I then rewrote that as (a+1)(a-1) < 0.
My second set of solutions then is a < -1 and a < 1.
two solutions: 0<a<1 or a<-1
Statement 1 is insufficient.

2) 1 - a^2 > 0 can be rewritten as (1+a)(1-a) >0. Your solutions are then a > -1 and a < 1.
two solutions a<-1 or a<1
Statement 2 is also insufficient.

combination
(0<a<1 or a<-1) and (a<-1 or a<1)
a<-1 is common in both

therefore we can answer the question: is a>0
ans = c

xiamsamx wrote:Is a>0?

(1) a^3 - a < 0
(2) 1 - a^2 > 0

Statement 1: a³ < a.
a³ < a.
a³ - a < 0.
a(a+1)(a-1) < 0.
The critical points are a=0. a=-1, a=1.
These are the only values for a 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:
Let a = -2.
(-2)³ < 2.
-8 < -2.
This works.

Plug -1 < a < 0 into a³ < a:
Let a = -1/2.
(-1/2)³ < -1/2
-1/8 < - 1/2.
Doesn't work.

Plug 0 < a < 1 into a³ < a:
Let a = 1/2.
(1/2)³ < 1/2
1/8 < 1/2.
This works.

Plug a > 1 into a³ < a:
Let a = 2
(2)³ < 2
8 < 2.
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 for a 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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Is a>0?

(1) a^3 - a < 0
(2) 1 - a^2 > 0

If the range of the question includes that of the condition, the condition is sufficient.
There is one variable (a), and so we need one equation, whereas 2 equations are given from the 2 conditions.

from condition 1, a(a-1)(a+1)<0 ==> a<-1, 0<a<1, but this is not included in the range of the question, so this condition is insufficient.
From condition 2, a^2-1<0, (a-1)(a+1)<0 ==> -1<a<1, and with the same reason this is also a insufficient condition.
But if we combine the 2 conditions, 0<a<1 is what we get, and this is included in the range of the question, so the conditions combined make a sufficient condition, and the answer is (C).

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