navalpike wrote:
Is X negative?
a) x^3(1-x^2)<0
b) x^2-1<0
First, Statement 2 gives us that x^2 < 1, or that -1 < x < 1, so is insufficient.
In Statement 1, we can divide both sides by x^2 (since x^2 must be positive, we don't need to be worried about whether to reverse the inequality) to get:
x*(1 - x^2)<0
or
x < x^3
This is an inequality comparing two different powers of x, and these show up often on the GMAT, so it's important to know how to analyse them (in fact, the inequality in Statement 2 is of the same type if you rewrite it as x^2 < 1; on one side we have x^2 and on the other we have x^0). First, notice that you can always multiply or divide both sides of these kinds of inequalities by any *even* power of x without needing to worry about reversing the inequality, since any even power of x cannot be negative. So, if you see:
x^97 < x^99
you can divide both sides by x^96 to get
x < x^3
and if you see
1/x^5 < 1/x^3
or equivalently
x^(-5) < x^(-3)
you can multiply on both sides by x^6 to get
x < x^3
Now, what does the inequality x < x^3 tell us? For which values of x is it true? We can break this down in a few ways. Of course you'll only need one method on the test, but I'll suggest three, each of which can be used reliably:
1. Notice that the left and right side of the inequality are equal when x = -1, 0 and 1. These three numbers are the only potential 'turning points' for the inequality; if you imagine gradually increasing x, for example, the inequality may change from being true to being false (or vice versa) as x passes -1, 0 and 1. This allows us to analyse the inequality by picking numbers:
-if the inequality is true for any number less than -1, it's true for all numbers less than -1;
-if the inequality is true for any number between -1 and 0, it's true for all numbers between -1 and 0;
-if the inequality is true for any number between 0 and 1, it's true for all numbers between 0 and 1;
-if the inequality is true for any number greater than 1, it's true for all numbers greater than 1.
So if you plug in -2, -1/2, 1/2 and 2, you can see exactly when the inequality is true or false. For the inequality x < x^3, we find that it's true whenever -1 < x < 0, and whenever x > 1.
This method tends to be very fast with some practice, and is what I'd normally use. You can only use this method when you have a simple inequality with different powers of x (or some other letter) on each side, but it can be used in this situation no matter what powers are involved.
There are algebraic methods as well:
2. Rewrite the inequality with 0 on one side, and a product on the other side. Here we have x - x^3 < 0, or x(1 - x^2) < 0. Since we are multiplying two things and getting a negative result, one of them must be positive, and one must be negative. That is:
*either x < 0, and 1 - x^2 > 0, which means that 1 > x^2, and -1 < x < 1. Combining that with the assumption that x < 0, we have that -1 < x < 0;
*or x > 0, and 1 - x^2 < 0, which means that x^2 > 1, and x > 1 or x < -1. Combining that with the assumption that x > 0, we have that x > 1.
So we find that either -1 < x < 0, or x > 1.
3. Rewrite the inequality with 0 on one side, and a complete factorization on the other:
x < x^3
0 < x^3 - x
0 < x(x^2 - 1)
0 < x*(x - 1)*(x + 1)
It's most useful to write these factors in increasing order:
0 < (x-1)*x*(x+1)
Now we have a product which is positive. Either:
-all three factors are positive, which means that even the smallest factor is positive, so x - 1 > 0, and x > 1.
-two factors are negative, and one is positive. The smaller two factors must be negative, so x < 0, and the largest factor must be positive, so x+1 > 0, or x > -1. Combining this information, we have that -1 < x < 0.
