Redhorsep wrote:Hi there,

I was stumped by this question which I found from OG 11th ed DS practice problems set second to last one:

Is x negative?

1. x^3 (1-x^2) < 0
2. x^2-1<0

I found OG's explanation to the first statement very wordy and confusing, can someone explain the solution with an emphasis on interpreting the first statement, will post the OG answer later. And also let me know how much time it takes for you to solve this, so I can have an idea how worthwhile it is to go through a similar one if it shows up on a real exam.

Thanks!

Is x negative?

1. x^3 (1-x^2) < 0
2. x^2-1<0

Explanation:

(1) x^3(1 - x²) < 0
If x = -2, then -8(1 - 4) = -8 * -3 = 24 > 0, so statement 1 is not true in case x is a negative number less than -1.
If x = -1/2, then (-1/8)(1 - 1/4) = (-1/8)(3/4) = -negative value < 0; statement 1 is true.
If x = 2, then 8(1 - 4) = -24 < 0; statement 1 holds true here.

From examples 2 and 3 taken above, it can be seen that x may or may not be a positive value; hence statement 1 is NOT sufficient.

(2) x² - 1 < 0 implies (x - 1)(x + 1) < 0 implies -1 < x < 1, which implies x may be a positive or a negative value; NOT sufficient again.

Combining (1) and (2), we know that -1 < x < 1, such that it lies between -1 and 0. Hence x is a negative value; SUFFICIENT.

The correct answer is C.