rxs0005 wrote:

Is x > x ^3 ?

(1) x < 0

(2) x^2 - x^3 > 2

You can either use inequalities here or the big picture approach.

When is x greater than x^3? It is when x < -1 or 0 < x <1.

(Recall that we should know the relations of x, x^2 and x^3 in the ranges 'less than -1', '-1 to 0', '0 to 1' and 'greater than 1')

(1) x < 0

When x is between -1 and 0, x^3 is greater than x. When x < -1, then x is greater than x^3. So this statement alone is not sufficient.

(2) x^2 - x^3 > 2

Now, this is not very easy to handle using inequalities. Without the cube, we would have just taken 2 to the other side and solved the quadratic. But this will be more easily solved using the big picture. Think of a number line.

What does x^2 - x^3 > 2 imply? It means x^2 is greater than x^3 and is 2 units to the right of x^3 on the number line. x^2 is never negative so it must be to the right of 0. Now there are two cases: Either x^2 is between 0 and 1 or it is greater than 1.

If x^2 were between 0 and 1, x would be between -1 and 1 and x^3 would be between 0 and -1. In this case, difference between x^3 and x^2 cannot be greater than 2. Hence this is not possible. So x^2 must be to the right of 1 and hence, x would be greater than 1 or less than -1. If x were greater than 1, x^3 would be greater than x^2 so x cannot be greater than 1. Hence x must be less than -1.

When x is less than -1, then x IS greater than x^3. Sufficient alone.

Answer (B)

Again, spend some time checking out the relations of x, x^2 and x^3 on the number line.

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