Inequality

This is a classic example for picking numbers:
(1) Insufficient: both a = 0.5 (or any other positive number less than 1) and b = -2 (or any other negative number less than -1) give the necessary result in the equation.

(2) Insufficient: both a = 0.5 (or any other positive number less than 1) and b = -0.5 (or any other negative number more than 1) give the necessary result in the equation.

However, when combining both of the statements we know that a must be a positive number less than 1.

Therefore, C

1. a^3 - a < 0 is equivalent to a(a^2 - 1) < 0 or P = a(a + 1)(a - 1) < 0. Now let's see when this happens:
a. first off, if a > 1, then P > 0. So we need to check cases before that.
b. when a < -1, then you have P = (negative numer)(negative number)(negative number), which is negative, so P < 0.
c. when a is between -1 and 0, then a + 1 turns positive, so P = (negative)(positive)(negative), which will be positive, so P > 0.
d. when a is between 0 and 1, then both a and a + 1 are positive, so P = (positive)(positive)(negative), which will be negative. So P < 0.
We are not checking the cases when a = -1, 0 or 1, since in any of these cases P = 0.

Now, as you can see, there are two cases in which P is negative:
- a < -1, when a < 0
- a is between 0 and 1, when a > 0.
So 1 is insufficient.

2. 1 - a^2 > 0 means that a^2 is smaller than 1. Since a^2 is greater than or equal to zero, then 0 <= a^2 < 1. This only happens when a is between -1 and 1. Again, you can't tell if a is positive, since -1 < a < 1.

Put both equations together and you get that the only section they intersect is when a is between 0 and 1, which means that a is positive.

My guess is C

Awesome solution Ron!

Regards,
CR

lunarpower wrote: of course, to get to this point in the first place, you have to determine that the individual statements are each insufficient; that's just routine factoring and solving. if anyone needs to know how to do that, post back.

Ron,
Can you explain how?

cunu-

Not sure if this detail will help you out, but I'll give it a shot.

Test: Is a>0?
(Supposition 1) a^3 - a < 0
I usually try to find the values where the function is equal to zero. If you know those values, all values in between will work out either < or > zero. For example, if two of these values are -1 and 1, you know that all the values less than -1 will be the same type of value (let's say < 0). It's probably easier to illustrate thus:

a^3 - a = 0
a (a^2 - 1) = 0 Simply factoring here
a (a - 1)(a + 1) = 0 Still factoring
a = 0; a = 1; a = -1 are the solutions when setting the function equal to zero.

From those solutions, we know that we have to test 4 different areas:
1. When a < -1
2. When -1 < a < 0
3. When 0 < a < 1
4. When 1 < a

1. Suppose a = -2; a (a - 1)(a + 1)
(-2)(-2 - 1)(-2 + 1) < 0
(-2)(-3)(-1) < 0
-6 < 0
True, so supposition (1) holds for a < -1

2. Suppose (-1/2); a (a - 1)(a + 1)
(-1/2)(-1/2 - 1)(-1/2 + 1) < 0
(-1/2)(-3/2)(1/2) < 0
(3/8) < 0
False, so (1) does not hold for -1 < a < 0

** At this point, since (Supposition 1) can be either greater than 0 or less than 0 (depending on chosen a), we know than (1) is insufficient for our Test. Although because it's pertinant to later discussion, I'll also solve for the other 2 areas (3. & 4. from above) **

3. Suppose (1/2); a (a - 1)(a + 1)
(1/2)(1/2 - 1)(1/2 + 1) < 0
(1/2)(-1/2)(3/2) < 0
(-3/8) < 0
True, so (1) holds for 0 < a < 1

4. Suppose a = 2; a (a - 1)(a + 1)
(2)(2 - 1)(2 + 1) < 0
(2)(1)(3) < 0
6 < 0
False, so (1) does not hold for 1 < a


(Supposition 2) 1 - a^2 > 0
Again, let's set it equal to zero to know which areas we need to test.
1 - (a)^2 = 0
(a)^2 - 1 = 0 Multiplied through by -1
(a - 1)(a + 1) = 0
a = 1 or a = -1

Since we have two values, we have three areas to test:
1. When a < -1
2. When -1 < a < 1
3. When 1 < a

1. Suppose a = -2; (a - 1)(a + 1)
(-2 - 1)(-2 + 1) < 0
(-3)(-1) < 0
3 < 0
False, so supposition (2) does not hold for a < -1.

2. Suppose 0; (a - 1)(a + 1)
(0 - 1)(0 + 1) < 0
(-1)(1) < 0
-1 < 0
True, so (2) does hold for -1 < a < 1.

** At this point, since (Supposition 2) can be either greater than 0 or less than 0 (depending on chosen a), we know than (2) is insufficient for our Test. Although because it's pertinant to later discussion, I'll also solve for the other 1 area (3. from above) **

3. Suppose 2; (a - 1)(a + 1)
(2 - 1)(2 + 1) < 0
(1)(3) < 0
3 < 0
False, so (2) does not hold for 1 < a.


Since we know that both suppositions and insufficient alone, we can now look at both together. If both together are true, where do their solutions overlap?

If a < -1, (1) is < 0 and (2) is > 0. This conflicts, so the solution to both cannot be in this area.

If a > -1, (1) is > 0 and (2) is < 0. This conflicts, so the solution to both cannot be in this area.

If -1 < a < 0, (1) is > 0 and (2) is < 0 (because we've seen that (2) is < 0 for all values between -1 and 1 non inclusive). This conflicts, so the solution to both cannot be in this area.

If 0 < a < 1, (1) is < 0 and (2) is < 0. This agrees, so the value if both suppositions are true must fall into this range. And since this solution allows us to answer our Test (a > 0?), we know that both (1) and (2) are sufficient for the Test.

Answer C.

PS: This is probably WAY more than you'll have time to do on the test, but it illustrates the steps in the most detail for your understanding. It is definitely not the most efficient solution.

BuckeyeT,
thanks alot. great explanation.

s a > 0?

1. a^3 - a < 0

(a-1) a (a+1) < 0

INSUF


2. 1 - a^2 > 0[/quote]

(1-a) ( 1+a ) > 0

INSUF

Together we know that (a+1)(a-1) < 0 SO a > 0

C