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x is a negative

by sanju09 » Sat Jun 19, 2010 12:17 am
If x is a negative number, which of the following must be true?
I. x^5 < |x|
II. x < √(-x)
III. x - 1/|x| < 0
(A) I, II and III
(B) I and II only
(C) II and III only
(D) I and III only
(E) None
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by amising6 » Sat Jun 19, 2010 12:32 am
sanju09 wrote:If x is a negative number, which of the following must be true?
I. x^5 < |x|
II. x < √(-x)
III. x - 1/|x| < 0
(A) I, II and III
(B) I and II only
(C) II and III only
(D) I and III only
(E) None

I. x^5 < |x| x is negative xo x^5 will also be neagtuve < |x| (ia always positive)
II) x < √(-x) not possible √(-x)

III. x - 1/|x| < 0 =(-x +1/x) <0 as 1/x is less than x and withnegative sign it will be less than 0

so option D
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by mj78ind » Sat Jun 19, 2010 1:31 am
I think A.

1 - for any value of negative x, x^5<abs(x)
2 - for any negative x, x is always less than sqrt (-x) since sqrt (-x) is positive
3 - for negative x, x - 1/abs(x) is always less than 0 since x and -1/ans(x) are both separately negative

Hence A
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by amising6 » Sat Jun 19, 2010 1:37 am
mj78ind wrote:I think A.

1 - for any value of negative x, x^5<abs(x)
2 - for any negative x, x is always less than sqrt (-x) since sqrt (-x) is positive (until and unless x is comples number this is not possible as squareroot of a negative number is not possible a gmat trap)
3 - for negative x, x - 1/abs(x) is always less than 0 since x and -1/ans(x) are both separately negative

Hence A
dude c the bold part
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by mj78ind » Sat Jun 19, 2010 1:43 am
amising6 wrote:
mj78ind wrote:I think A.

1 - for any value of negative x, x^5<abs(x)
2 - for any negative x, x is always less than sqrt (-x) since sqrt (-x) is positive (until and unless x is comples number this is not possible as squareroot of a negative number is not possible a gmat trap)
3 - for negative x, x - 1/abs(x) is always less than 0 since x and -1/ans(x) are both separately negative

Hence A
dude c the bold part
@amising6
Nice observation the only issue is it is wrong. Say x = -6, then if you plug it in stmt 2 above we get -6 < Sqrt{-(-6)}

which boils down to -6 < sqrt(6), hence is always true and x does not have to be complex.

Cheers
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by amising6 » Sat Jun 19, 2010 1:55 am
yes agrreod oversight error i ignore that part
thanks
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