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first we know that X is negative and it's between -1 and 0,

1)x3<x2-correct because x3 is negative(---=-) and x2 is posittive(--=+) 2)x5<1-x x5 is negative and 1 minus negative gives 1+positive so it's correct 3)x4<x2 here u should remember that if number is between 0 and 1, square gives less than actual number...u can check for example 0.5 in square is 0.25 which is less than 0.5..so X4 is less than x2
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As X^2 will always be positive or zero. only (X-1) contributes to the negativity of the above inequality.

(X-1) < 0 => X <1 so it must be true for X to be in range (-1,0)

(2). X^5 < 1 -X

=> X(X^4+1) < 1 => X <1 Must be true (As X^4 +1 wil always be greater than 1)

(3). X^4 < X^2

=>X^2(X^2-1) <0 => (X+1)(X-1) <0

=> -1<X<1 MUST be true

Hence , (E) !!
_________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

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

I. x^3 < x^2 --> from -1 < x < 0 it follows that LHS<0<RHS, so this statement is true.

II. x^5 < 1 – x --> x(x^4+1) < 1 --> negative*positive < 0 < 1, so this statement is also true.

III. x^4 < x^2 --> reduce by x^2 (we can safely do that since from -1 < x < 0 is follows that x^2>0): x^2 < 1. Again, as -1 < x < 0, then x^2 must be less than 1. Hence, this statement is also true.

Fractions get smaller every time they are multiplied together. For negative numbers raised to an odd power the result will be negative. Knowing these two things makes the comparisons fairly simple.
_________________

I do not beg for kudos.

gmatclubot

Re: If -1 < x < 0, which of the following must be true?
[#permalink]
27 Jul 2013, 11:25

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