Problem Solving

If -1 < x < 1 and x ≠ 0, which of the following inequalities must be true?
I. x3 < x
II. x2 < |x|
III. x4 - x5 > x3 - x2
(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III


OA D

I an getting option II as wrong.

Here is what I did,

Option II: x2 < |x|

Let's take X = 0.1

(0.1)^2 < 0.1 ==> YES

Now, X = -0.1

(-0.1)^2 < -0.1 ==> NO
Hence Option II is insufficient.

[email protected] wrote:Hi Uva@90,

In your second example, you forgot to include the absolute value symbol. It should be....

(-.1)^2 < |-.1|
.01 < .1

This is TRUE.

GMAT assassins aren't born, they're made,
Rich

No RICH,

I didn't forget,
|X| = -x when x <= 0 right ?

Regards,
Uva

I didn't forget,
|X| = -x when x <= 0 right ?

Yes, but if x = -.1 then -x = -(-.1) = .1
(Or, put another way, when x is negative, (-x) will be positive.)

Uva@90 wrote:If -1 < x < 1 and x ≠ 0, which of the following inequalities must be true?
I. x³ < x
II. x² < |x|
III. x� - x� > x³ - x²

(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III

Since -1 < x < 1 and x ≠ 0, x must a NEGATIVE OR POSITIVE FRACTION.

Statement I: x³ < x
If x = -1/2, then x³ = -1/8.
In this case, x³ > x.
Since it does not have to be true that x³ < x, eliminate A and E.

Statement II: x² < |x|
Since x is nonzero, x² > 0 and |x| > 0.
Since both sides of the inequality are positive, we can square the inequality:
(x²)² < (|x|)²
x� < x².

Since x² > 0, we can divide both sides by x²:
x�/x² < x²/x²
x² < 1.

Since the square of a negative or positive fraction must be less than 1, statement II must be true.
Eliminate C.

Statement III: x� - x� < x² - x³
Since x is nonzero, we can divide by x², which must be a positive value:
(x� - x�)/x² < (x² - x³)/x²
x² - x³ < 1-x
x²(1-x) < 1-x

Since x is a negative or positive fraction, we can divide by 1-x, which also must be a positive value:
x²(1-x)/(1-x) < (1-x)/(1-x)
x² < 1.

Since the square of a negative or positive fraction must be less than 1, statement III must be true.
Eliminate B.

The correct answer is D.