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If -1 < x < 1 and x!=0, which of the following inequal

Expert replies
by alanforde800Maximus » Mon Oct 10, 2016 8:02 pm
If -1 < x < 1 and x!=0, which of the following inequalities must be true ?

I. x^3 < x

II. x^2 < |x|

III. x^4 - x^5 > x^3 - x^2


a) I only

b) II only

c) III only

d) II and III only

e) I, II and III.

Please assist with above problem.
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Source: — Problem Solving |

by GMATGuruNY » Mon Oct 10, 2016 11:58 pm
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.
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by DavidG@VeritasPrep » Tue Oct 11, 2016 10:19 am
alanforde800Maximus wrote:If -1 < x < 1 and x!=0, which of the following inequalities must be true ?

I. x^3 < x

II. x^2 < |x|

III. x^4 - x^5 > x^3 - x^2


a) I only

b) II only

c) III only

d) II and III only

e) I, II and III.

Please assist with above problem.
Another way to think about statement III:

x^4 - x^5 > x^3 - x^2
Add x^5 and x^2 to both sides to get x^4 + x^2 > x^5 + x^3.
Factor out x^2 on the left and x^3 on the right to get x^2 * (x^2 + 1) > x^3 * (x^2 +1)
Divide both sides by x^2 + 1 (we know this term must be positive, so we don't have to worry about the inequality sign flipping) to get x^2 > x^3
Divide both sides by x^2 to get 1>x. So we know this is true.
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by [email protected] » Tue Oct 11, 2016 10:42 am
Hi alanforde800Maximus,

Most Roman Numeral questions on the Official GMAT are based on Number Properties and this question is no exception. You might find that TESTing VALUES helps you to better understand the patterns involved here...

We're told that -1 < X < 1 and that X ≠ 0. We're asked which of the following MUST be true.

RN 1: X^3 < X

IF...X = 1/2....1/8 < 1/2
IF...X = -1/2....-1/8 is NOT < -1/2
Roman Numeral 1 is NOT always TRUE.
Eliminate Answers A and E

RN 2: X^2 < |X|

Since X is either a positive fraction or a negative fraction, X^2 will be 'closer to 0' than X (so X^2 will be LESS than |X|.

For example....
IF...X = 1/2....1/4 < |1/2|
IF...X = -1/2....1/4 < |-1/2|
Roman Numeral 2 IS ALWAYS TRUE
Eliminate Answer C

RN 3: X^4 - X^5 > X^3 - X^2

This one is complex-"looking", but if you pay attention to the exponents it's not actually that tough.

IF...X = positive....
X^4 > X^5 and X^2 > X^3

So....
X^4 - X^5 = positive
X^3 - X^2 = negative
a positive > a negative, so this option is ALWAYS TRUE

IF....X = negative
X^4 and X^2 = positive
X^5 and X^3 = negative

X^4 - X^5 = (+) - (-) = positive
X^3 - X^2 = (-) - (+) = negative
Again, a positive > a negative, so this option is ALWAYS TRUE
Roman Numeral 3 IS ALWAYS TRUE
Eliminate Answer B.

Final Answer: D

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