If -1 < x < 0, which of the following must be true?

guerrero wrote:If -1 < x < 0, which of the following must be true?

I. x^3 < x^2
II. x^5 < 1 – x
III. x^4 < x^2

A. I only
B. I and II only
C. II and III only
D. I and III only
E. I, II and III

OAE

Constraint: -1 < x < 0.
Thus, x is a NEGATIVE FRACTION between -1 and 0.

I: x³ < x²
Since x≠0, x²>0.
Thus, we can safely divide both sides by x²:
x³/x² < x²/x²
x < 1.
Since x is a negative fraction between -1 and 0, it must be true that x<1.
Eliminate C, which does not include statement I.

III: x� < x²
Since x≠0, x²>0.
Thus, we can safely divide both sides by x²:
x�/x² < x²/x²
x² < 1.
Since x is a negative fraction between -1 and 0, it must be true that x² < 1.
Eliminate A and B, which do not include statement III.

II: x� < 1-x
x� + x < 1.
Since x<0, x� < 0.
Thus, x� + x = negative + negative = negative.
Thus, it must be true that x� + x < 1.
Eliminate D, which does not include statement II.

The correct answer is E.

One approach is plugging in a value of x (say, x = -1/2) and seeing which inequalities are true.

Another is to do it algebraically.

I:

x³ < x²

Since x² must be positive (as x is not 0), dividing both sides by x² won't flip the inequality. When we divide by x², we get

x < 1

Since x is negative, this is true, and I works.

II:

x� + x < 1
x(x� + 1) < 1

Since x is negative, and (x� + 1) is positive, x(x� + 1) must be negative, and hence < 1. II is true!

III:

x� < x²

0 < x² - x�

0 < x²(1 - x²)

Since x² is positive and (1 - x²) is positive (as 0 < x² < (-1)²), this is also true. III works.